Engineering.mechanics.dynamic.bedford.ch18

Problem 18.1

A horizontal force F = 30 lb is applied to the 230-lb refrigerator as shown. Friction is negligible. (a) (b)

What is the magnitude of the refrigerator’s acceleration? What normal forces are exerted on the refrigerator by the floor at A and B ?

F

60 in

28 in

A

14 in

Solution: Assume that the refrigerator rolls without tipping. We have the following equations of motion.



230 lb



Fx : ( 30 lb ) =

 

Fy : A + B − 230 lb = 0

32.2 ft/s2



F

=

14 in

30 lb 230 lb

a 60 in.

MG : −(30 lb )(32 in. ) − A(14 in. ) + B( 14 in. ) = 0

Solving we find 28 in.

(a)

a = 4.2 ft/s2

(b)

A = 80.7 lb, B = 149.3 lb

A

B

Since A > 0 and B > 0 then our assumption is correct.

c 2005 Pearson Educat ion, Inc., Upper Saddle River, NJ. All rights reserve d. This material is protected under all copyright laws as they  currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Problem 18.2

Solve Problem 18.1 if the coefficient of kinetic friction at A and B is µk = 0.1. Solution: Assume sliding without tipping



Fx : ( 30 lb ) − (0.1)(A + B) =

 

Fy : A + B − 230 lb = 0



230 lb 32.2 ft/s2



F

=

30 lb

a 230 lb 60 in.

MG : −(30 lb )(32 in. ) − A(14 in. ) + B( 14 in. ) 28 in.

− (0.1)(A + B)( 28 in.) = 0

µA

Solving, we find

µB N

(a)

a = 0.98 ft/s2

(b)

A = 57.7 lb, B = 172 lb

A

B


1

Problem 18.3

As the airplane begins its takeof f run, the normal forces exerted on its tires by the runway at A and B are NA = 720 lb and NB = 1660 lb. What is the magnitude of the airplane’s acceleration?

T

4 ft

3 ft W A

B

5 ft

Solution: We are given that N A = 720 lb, NB = 1660 lb. We use g = 32.2 ft/s2

2 ft

W

T

There are three unknowns (the weight W , the thrust T , and the acceleration a ).



Fx : −T =



Fy : N A + N B − W = 0



MG : T (1 ft ) − NA (5 ft ) + NB (2 ft ) = 0

W g

a

NA

NB

Solving we find T = 280 lb, W = 2380 lb, a = 3.79 ft/s2 a = 3.79 ft/s2

c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they  currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Problem 18.4

The mass of the Boeing 777 is 300,000 kg. As it begins its takeoff run, its two engines exert a total horizontal thrust T = 670 kN. Neglect horizontal forces exerted on the tires. (a) (b)

What is the magnitude of the airplane’s acceleration? What normal forces are ex erted on the tire s at A and B ?

T

5m

3m A

26 m

Solution: We are given

B 2m

mg

m = 300,000 kg, T = 670 kN, g = 9.81 m/s2



Fx : −T = −ma



Fy : N A + NB − mg = 0



MG : −T (2 m) − NA (24 m) + NB (2 m) = 0

T

NA

NB

Solving we find (a)

a = 2.23 m/s2

(b)

NA = 175 kN , NB = 2770 kN


1

Problem 18.5

The crane moves to the right with constant acceleration, and the 800-kg load moves without swinging. (a) (b)

What is the acceleration of the crane and load? What are the ten sions in the cabl es attached at A and B ? 5°

5°

A

B

1m

1.5 m

Solution: From Newton’s second law: F x = 800a N. The sum of the forces on the load:



◦



◦ ◦ Fy = FA cos 5 + FB cos 5 − 800g = 0.

The sum of the moments about the center of mass:



◦

5˚

5˚

FA

◦

Fx = FA si n 5 + FB sin 5 − 800a = 0.

1.5 m

FB

1.0 m

mg 1.5 m

1.5 m

◦

MCM = −1.5FA co s 5 + 1.5FB cos 5 ◦

◦

− FA si n 5 − FB si n 5 = 0.

Solve these three simultaneous equations: a = 0.858 m/s2 ,

FA = 3709 N ,

FB = 4169 N


1

Problem 18.6

The total weight of the go-cart and driver is 240 lb. The location of their combined center of mass is shown. The rear drive wheels together exert a 24lb horizontal force on the track. Neglect the horizontal forces exerted on the front wheels. (a) (b)

What is the magnitude of the go-cart’s acceleration? What normal forces are ex erted on the tire s at A and B ?

15 in 6 in

4 in

A

B 16 in 60 in

Solution: Fx : ( 24 lb ) =

  



240 lb 32.2 ft/s2



a

Fy : N A + NB − (240 lb) = 0 240 lb

MG : −NA (16 in. ) + NB (44 in. ) + (24 lb )(15 in.) = 0

Solving we find 24 lb

(a)

a = 3.22 ft/s2

(b)

NA = 182 lb , NB = 58 lb

NA

NB


1

Problem 18.7 The total weight of the bicycle and rider is 160 lb. The location of their combined center of mass is shown. The dimensions shown are b = 21 in. , c = 16 in., and h = 38 in. What is the largest acceleration the bicycle can have without the front wheel leaving the ground? Neglect the horizontal force exerted on the front wheel by the road. Strategy: You want to determine the value of the acceleration that causes the normal force exerted on the front wheel by the road to equal zero. h

A

B b

c

Solution: Given: b = 21 in., c = 16 in., h = 38 in.

160 lb

Find: a so that N A = 0



160 lb





Fx : −FB = −

 

Fy : NA + NB − (160 lb) = 0

32.2 ft/s2

a

MG : −NA b + NB c − FB h = 0 h

NA = 0

Solving we find N B = 160 lb, FB = 67.4 lb,

a = 13.6 ft/s

2

b

N

c

FB

NB

A


1

Problem 18.8

The total mass of the bicycle and rider is 72 kg. The location of their combined center of mass is shown. The dimensions shown are b = 530 mm , c = 400 mm, and h = 960 mm. If the bicycle is traveling at 5 m/s and the rider engag es the brakes, achieving the largest deceleration for which the rear wheel will not leave the ground, how long does it take the bicycle to stop, and what distance does it travel during that time? Solution: Given: mg

m = 72 kg , b = 530 mm c = 400 mm, h = 960 mm

Find: a so that N B = 0



Fx : F A = −ma



Fy : NA + NB − mg = 0



MG : −NA b + NB c + FA h = 0

h

FA

NB = 0

b

Solving we find N A = 706 N , FA = 390 N , a = −5.42 m/s2 NA

Now do the kinematics

c

NB

a = −5.42 m/s2 v = (−5.42 m/s2 )t + (5 m/s ) s = (−2.71 m/s2 )t 2 + (5 m/s )t

When the bicycle stops v = (−5.42 m/s2 )t + (5 m/s ) = 0 ⇒ t = 0.923 s

The bicycle has traveled 2

2

s = (−2.71 m/s )t + (5 m/s )t ⇒ s = 2.31 m


1

Problem 18.9

The combined mass of the motorcycle and rider is 160 kg. The rear wheel exerts a 400-N horizontal force on the road, and you can neglect the horizontal force exerted by the front wheel. Modeling the motor cycle and its wheels as a rigid body, determine (a) the motorcycle’s acceleration and (b) the normal forc es exerted on the road by the rear and front wheels.

660 mm

A

B

660 mm 1500 mm

The friction force on the rear wheel is F = 400 N. From Solution: Newton’s second law, the acceleration is

a=

F m

=

400 160

= 2.5 m/s 2 .

The sum of the forces in the y direction is



Fy = NA + NB − mg = 0.

The moment about the center of mass is



F

mg NA

NB

MCM = −0.66NA + (1.5 − 0.66)NB + 0.66(400) = 0.

Solve these two equations in two unknowns: NA = 1055 N ,

NB = 514.6 N


1

Problem 18.10

The moment of inertia of the disk about O is 22 kg-m 2 . At t = 0, the stationary disk is subjected to a constant 50 N-m torque. (a) (b)

What is the magnitude of the disk’s angular velocity at t = 5 s? Through how many revolutions does the disk rotate from t = 0 to t = 5 s?

50 N-m

O

Solution: The dynamics



MO : ( 50 Nm ) = (22 kg-m2 )α ⇒ α = 2.27 rad/s2

Kinematics α = 2.27 m/s2 , ω = (2.27 m/s2 )t, θ = (1.14 m/s2 )t 2

At t = 5 s, (a)

ω = 11.4 rad/s

(b)

θ = 28.4 rad



1 rev 2π rad



= 4.52 rev


1

Problem 18.11

During extravehicular activity, an astronaut fires a thruster of his maneuvering unit, exerting a force T = 14.2 N for 1 s. It requires 60 s from the time the thruster is fired for him to rotate through one revolution. If you model the astronaut and maneuvering unit as a rigid body, what is the moment of inertia about their center of mass? T

300 mm

Solution: The angular velocity is

ω = αt =

  2π 60

rad/s.

Since the thruster fires for one second, α=

2π

rad/s2 .

60 The sum of the moments:



MCM = T r = 14.2(0.3) = 4.26 N-m .

From the equation of angular motion Iα=



MCM = 4.26 N-m,

from which

I =

4.26 2π

 

= 40.7 kg-m 2

60


1

Problem 18.12

The moment of inertia of the helicopter’s rotor is 420 slug-ft 2 . The rotor starts from rest at t = 0. The torque exerted on it by the engine is given as a function of time by 500 − 20t ft-lb. (a) (b)

What is the magnitude of the rotor’s angular velocity at t = 10 s? Through how many revolutions does the rotor rotate from t = 0 to t = 10 s?

Solution: The dynamics and kinematics:



M : ( 500 ft-lb − [20 ft-lb/s]t) = (420 lb-ft-s2 )α

α = (1.19 rad/s2 ) − (0.0476 rad/s3 )t ω = (1.19 rad/s2 )t − (0.0238 rad/s3 )t 2 θ = (0.595 rad/s2 )t 2 − (0.00794 rad/s3 )t 3

At t = 10 s (a)

ω = 9.52 rad/s

(b)

θ = 51.6 rad



1 rev 2π rad



= 8.21 rev


1

Problem 18.13

The mom ent of inertia of the robot manipulator arm about the vertical y axis is 10 kg-m 2 . The mome nt of inertia of the 14-k g casting held by the arm about the y  axis is 1 .2 kg-m 2 . The system is initially stationary. At t = 0, the arm is subjected to a torque about the y axis that is given as a function of time by 220 + 100t N-m. How long does it take the arm to undergo one revolution?

y

y'

Manipulator arm

Casting 0.8 m

Solution: The moment of inertia of the structur e about the axis is

y-

Iy = (10 kg-m2 ) + (1.2 kg-m2 ) + (14 kg )(0.8 m )2 = 20.16 kg-m2

 α=

My : ( 220 N-m) + (100 N-m/s)t = (20.16 kg-m2 )α



220 20.16



rad/s2 +



100 20.16





rad/s3 t

Thus α = (10.91 rad/s2 ) + (4.96 rad/s3 )t ω = (10.91 rad/s2 )t + (2.48 rad/s3 )t 2 θ = (5.46 rad/s2 )t 2 + (0.827 rad/s3 )t 3

We set θ = 1 rev = 2π rad and solve to find

t = 1.00 s

c 2005 Pearson Educat ion, Inc., Upper Saddle River, NJ. All rights reserve d. This material is protected under all copyright laws as they  currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Problem 18.14 The mom ent of inertia of the windtunnel fan is 225 kg-m 2 . The fan starts from rest. The torque exerted on it by the engine is given as a function of the angular velocity of the fan by T = 140 − 0.02ω2 N-m.

(a)

When the fan has turned 620 revolutions, what is its angular velocity in rpm (revolutions per minute)? What maximum angular velocity in rpm does the fan attain?

(b)

Strategy: By writing the equation of angula r motion, determine the angular acceleration of the fan in terms of its angular velocity. Then use the chain rule: α=

dω dt

=

dω dθ dθ dt

=

dω dθ

ω.

Solution:

 α=

M : ( 140 N-m) − (0.02 N-m/s2 )ω2 = (225 kg-m2 )α



140 225

rad/s2

  −

0.02 225



rad/s4 ω2

= (0.622 rad/s2 ) − (0.0000889 rad/s4 )ω2

(a)

α=ω



dω dθ

= (0.622 rad/s2 ) − (0.0000889 rad/s4 )ω2

ω

0

ωd ω (0.622 rad/s2 ) − (0.0000889 rad/s4 )ω2

=



620(2π) rad

dθ 0

Solving we find

ω = 59.1 rad/s

(b)



1 rev 2π rad



60 s 1 min



= 565 rpm

The maximum angular velocity occurs when the angular acceleration is zero α = (0.622 rad/s2 ) − (0.0000889 rad/s4 )ω2 = 0

ω = 83.7 rad/s



1 rev 2π rad



60 s 1 min



= 799 rpm


1

The gears A and B can turn freely on their pin supports. Their moments of inertia are IA = 2 0.002 kg-m and IB = 0.006 kg-m 2 . They are initially stationary, and at t = 0 a constant couple M = 2 N-m is applied to gear B . How many revolutions has gear A turned at t = 4 s? Problem 18.15

M

90 mm B A

60 mm

Solution: Use the usual conventions of directions for angles and angular accelerations. The radius of gear A is 60 mm, and the radius of gear B is 90 mm. The tangential accelerations at the point of contact are equal: − rA αA = rB αB . From the equation of angular motion

M

B F A

M − F rB = IB αB ,

F

and −F rA = IA αA ,

from which M+

  rB

rA

2=−

αA =

IA αA = IB αB

  rA

rB

IB αA −

  rB

rA

−2

  rA

rB

IB +

  rB

rA

IA αA ,

= −285.7 rad/s2

IA

The angle of revolution in 4 seconds is αA n=

2

  2

(4 ) = −2285.7 rad,

from which N=

n

2π

= −363.8 revs


1

Problem 18.16

The disks A, B , and C do not slip relative to each other at their points of contact. Their masses are mA = 4 kg, mB = 16 kg, and mC = 9 kg. They are initially stationary. At t = 0, a constant 10 N-m counterclockwise couple is applied to disk A. What is the angular velocity of disk C at t = 5 s?

10 N-m 0.4 m 0.3 m

0.2 m

A

Solution: The FBDs

10 N−m

B

C

F1 F2

The dynamics



MA : ( 10 N-m ) − F1 (0.2 m ) =



MB : −F1 (0.4 m ) + F2 (0.4 m ) =



1 2

(4 kg)(0.2 m )2 αA B

MC : F 2 (0.3 m ) =

1 2

1 (16 kg )(0.4 m )2 αB 2

A

F1

C

F2

2

(9 kg)(0.3 m ) αC

The kinematic constraints αA (0.2 m ) = −αB (0.4 m ), −αB (0.4 m ) = αC (0.3 m )

Solving five equations for five unknowns we find F1 = 43.1 N, F2 = 15.5 N αA = 17.2 rad/s2 , αB = −8.62 rad/s2 , αC = 11.5 rad/s2

Kinematics for gear C αC = 11.5 rad/s2 , ω = (11.5 rad/s2 )t, θ = (5.75 rad/s2 )t 2

At t = 5 s we have

ω = 57.5 rad/s

CC W


1

Problem 18.17

The moment of inertia of the pulley is 0.4 slug-ft2 . The 5-lb weight slides on the smooth horizontal surface. If the system starts from rest, determine how far to the right the 5-lb weight moves in 1 s in each case.

5 lb

5 lb

6 in

6 in 2 lb

2 lb (a)

(b)

Solution: 5 lb

(a)

The dynamics: T



5 lb 32.2 ft/s2

6

M:T





Fx : T =



a

6

ft − (2 lb)

12

The kinematic constraint:

= −(0.4 lb-s2 -ft)α

ft

N

12

  a=

    6

12

2 lb

ft α 5 lb

Solving we find 2

2

a = 1.14 ft/s , T = 0.177 lb, α = 2.28 rad/s

T1

The kinematics: a = 1.14 ft/s2 , v = (1.14 ft/s2 )t, s = (0.570 ft/s2 )t 2

At t = 1 s we have (b)

s = 0.570 ft

N T2

The dynamics



5 lb



Fx : T 1 =



M : T1



Fy : T 2 − (2 lb) = −

32.2 ft/s2



a

    6

12

6

ft − T2

The kinematic constraint:

12

ft

= −(0.4 lb-s2 -ft)α



2 lb 32.2 ft/s2

a=

  6

12

2 lb



a

ft α

Solving we find a = 1.10 ft/s2 , T1 = 0.171 lb, T2 = 1.93 lb, α = 2.20 rad/s2

The kinematics: a = 1.10 ft/s2 , v = (1.10 ft/s2 )t, s = (0.550 ft/s2 )t 2

At t = 1 s we have

s = 0.550 ft


1

Problem 18.18

The masses of the slender bar and disk are 5 kg and 10 kg, respectively. The coefficient of kinetic friction between the disk and the horizontal surface is µk = 0.1. Determine the disk’s angular acceleration if it is subjected to (a) an 8 N-m counterclockwise couple; (b) an 8 N-m clockwise couple.

30°

1.2 m

0.4 m

Solution:

d = 0.6cos30 ◦

Ay

d

d

h = 0.6sin30 ◦ Ax

30

A

mB = 5 kg

°

h

mD = 10 kg ID =

h

1 m R2 2 D

Bx

mBg

R = 0.4 m

By

ID = 0.8 kg-m2 µk = 0.1 Mc

Mc = 8 N-m

mDg

Bar:



Ax + Bx = 0



Ay + By − mB g = 0

Fx : Fy :



MA :

Bx

µ k ND

− dm B g + 2dBy + 2hBx = 0

Disk:

ND



− Bx − µk ND = 0



− By − mD g + ND = 0



Mc − Rµk ND = ID α

Fx : Fy : MB :

By

6 eqns in 6 unknowns

See solution to Problem 18.18(a). There are two changes. The value for Mc becomes − 8 N-m and the direction of the friction force changes. The equations for the disk become



− Bx + µk ND = 0



− By − mD g + ND = 0



Mc + Rµk ND = ID α

Fx : Fy :

(Ax , Ay , Bx , By , ND , α)

MB :

Solving,

Ax = 13.01 N, Bx = −13.01 N,

Ay = 17.01 N By = 32.04 N

(changed)

(unchanged)

(changed)

Note — the sig n on Mc in the equation did not changed because we changed the numerical value. Solving the six eqns, we get

Ax = −11.59 N,

Ay = 31.22 N

ND = 130.14 N Bx = 11.59 N,

By = 17.83 N

2

α = 3.49 rad/s (counterclockwise)

ND = 115.9 N α = −4.20 rad/s2

(clockwise)


1

Problem 18.19 The 5-kg slender bar is released from rest in the horizontal position shown. Determine the magnitude of the bar’s angular velocity when it has fallen to the vertical position.

1.2 m

Strategy: By drawing a free-body diagram of the bar when it has fallen through an arbitrary angle θ and using the equation of angular motion, determine the bar’s angular acceleration as a function of θ . Then apply the chain rule: α=

dω dt

=

dω dθ dθ dt

=

dω dθ

ω.

Solution:



α=ω



dω dθ

ωd ω =

2

3

m(1.2 m )2 α

2



π/ 2

Ox θ

(24.5 rad/s2 ) sin θ d θ

0

mg π/ 2

= −(12.3 rad/s2 )[cos θ ]0

Thus

1

= (12.3 rad/s ) sin θ

ω

0

ω2

Oy

MO : m( 9.81 m/s2 )(0.6 m ) sin θ =

= 12.3 rad/s2

ω = 4.95 rad/s


1

Problem 18.20 The objects cons ist of identical 3-ft, 10-lb bars welded together. If they are released from rest in the positions shown, what are their angular accelerations and what are the components of the reactions at A at that instant? (The y axes are vertical.)

y

y

x

A

45

°

A

(a)

x

(b)

Solution: The angular velocity ω = 0 at the inst ant of interest since the system is released from rest. The moment of inertia of a slender rod about the end is Iend = mL2 /3. The moment of inertia about the center of the slender rod is Icenter =

  1

12

Ay

mL2 .

Ax 2Wb

(a) The moment of inertia of system (a) about



IA = mL2 1 +



=

10 32.17

1 3

+

1 12

 

XC

12 + 4 + 1

(32 )

A is

 12



= 3.96 slug-ft2 .

The center of mass of the system is

m xC =

The reaction at A : A x = 0,

Ay = 2Wb +

 

  2Wb g

ay = 20 +

20 32.17

(−25.5) = 4.12 lb .

L

+ mL 2 = 2.25 ft . 2m

(b) The moment of inertia of system (b) about IA = mL2

The moment about A is

M = rC/A × −2mgj =

 

i xC 0

j yC −2mg

k 0 0

 

=





1+

10 32.17

1 3

1

+

12

  (32 )

A is



12 + 4 + 1 12



= 3.96 slug-ft2 .

The center of mass of the system is

= −2mgxC k = 45k ft-lb.

m(L/2) + mL

From the equation of angular motion,

xC =

M = IA α,

These are identical with the results of part (a), from which

from which α =

M IA

=

45

= 11.4 rad/s2 .

3.96

The reactions at A are obtained from Newton’s second law: Ay − 2Wb =

2Wb g

ay ,

2Wb

Ax =

g

ax ,

2m

= 2.25 ft .

α = 11.4 rad/s2 ,

Ay = 2Wb + (2Wb /g)ay = 4.12 lb , and

Ax = 0

where a x , a y are the accelerations at the center of mass of the system. From the kinematic relationships: aCM = α × rC/A − ω2 rC/A

from which, since ω = 0,

aCM = α × rC/A =

 

i 0 −xC

j 0 0 2

= −αxC j = −25.55j ft/s .

k α 0

 


1

Problem 18.21

The object consists of the 2-kg slender bar ABC welded to the 3-kg slender bar BDE. The y axis is vertical. The object is released from rest in the position shown. Determine its angular acceleration and the components of the force exerted on the object by the pin at D at the instant it is released.

A

y

0.2 m

B

D

E

x 0.4m

0.2 m

0.2m

C

Solution: The center of mass is located to the left of poin t D a distance d=

(3 kg)(0.1 m ) + (2 kg)(0.4 m )

5 kg

Dy

= 0.22 m .

The moment of inertia about point O is ID =

1 12

2

Dx 2

(2 kg)(0.4 m ) + (2 kg)(0.4 m )

(5 kg)(9.81 m/s2)

1 + (3 kg)(0.6 m )2 + (3 kg)(0.1 m )2 = 0.467 kg-m2 12 The FBD Dynamics equations



MD : ( 5 kg)(9.81 m/s2 )(0.22 m ) = (0.467 kg-m2 )α



Fx : D x = 0



Fy : D y − (5 kg)(9.81 m/s2 ) = −(5 kg)a

Kinematic constraint:

a = (0.22 m )α

Solving we find Dx = 0, Dy = 23.6 N , α = 23.1 rad/s

CC W


1

Problem 18.22

Determine the angular acceleration and the compo nents of the forc e exerted on the objec t in Problem 18.21 by the pin at D at the inst ant the bar BDE has fallen to the vertical position.



Solution: At an arbitrary angle θ we have

Dx

α=ω



dω dθ

ωd ω =

ω2

2

2

= (23.1 rad/s ) sin θ

ω

0



π/ 2

(23.1 rad/s2 ) sin θ d θ

0

π/ 2

= −(23.1 rad/s2 )[cos θ ]0

= 23.1 rad/s2

ω 2 = 46.2 s −2

  

Dy

MD : ( 5 kg)(9.81 m/s2 )(0.22 m ) sin θ = (0.467 kg-m2 )α

(5 kg)(9.81 m/s2)

In the vertical position we have the dynamic equations MD : 0 = (0.467 kg-m2 )α Fy : D y − (5 kg)(9.81 m/s2 ) = may

Fx : D x = max

and the kinematic constraints ax = (0.22 m )α ay = (0.22 m )ω 2

Solving we find

ax = α = Dx = 0, Dy = 99.9 N


1

Problem 18.23

For what value of x is the horizontal bar’s angular acceleration a maximum, and what is the maximum angular acceleration?

m

x l

1 I mx 2 mL2 . The 12 moment about the pin is M mgx . From the equation of angular motion, I α mgx, from which Solution: The mome nt of inertia is

=

=

α

= gx



2

x2

+ L12

=

+

1

−

.

Take the derivative: dα dx

2

+ L12



Solve:

x

x2

=0=g

= √L

12

,

−1



2

1

− 2x 2

 and

αmax

x2

+ L12

 =

−1



.



√

3g

L

c 2005 Pearson Educat ion, Inc., Upper Saddle

River, NJ. All rights reserve d. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Problem 18.24 Model the arm ABC as a single rigid body. Its mass is 300 kg, and the mome nt of inertia about its center of mass is I = 360 kg-m 2 . If point A is stationary and the angular acceleration of the arm is 0.6 rad/s 2 counterclockwise, what force does the hydraulic cylinder exert on the arm at B ? (The arm has two hydraulic cylinders, one on each side of the vehicle. You are to determine the total force exerted by the two cylinders.)

y 1.80 m

C

1.40 m

B

0.30 m 0.80 m

x

A 0.70 m

Solution: The angle of the hydraulic cylinder with the horizontal is θ = tan−1

  1.5 1.4

◦ = 47 .

Ay mg

The moment of inertia about A is F IA = 360 + (1.82 + 1.12 )(300) = 1695 kg-m2 .

Ax

The moment is the sum of the moments exerted by the cylinder and the weight of the arm:

θ

M = rAB × F + rCM × mg.

M=

 

i 1.4 F cos θ

j 0.8 F sin θ

k 0 0

    +

i 1.8 0

j 1.1 −mg

k 0 0

 

M = (0.4776F − 5297.4)k (N-m).

From the equation of angular motion: M = IA α , from which (0.4776F − 5297.4)k = 1695(0.6)k, and F = 13220 N


1

Problem 18.25 The mass of the truck’s bed is 2500 kg and its moment of inertia about O is 78 ,000 kg-m 2 . At the instant shown, the coordinates of the center of mass of the bed are (3, 3.75) m and the coordinates of point B are (4.5, 3.5) m. If the angul ar acceleration of the bed is 0 .5 rad/s 2 in the clockwise direction, what is the magnitude of the force exerted on the bed at B by the hydraulic cylinder AB ?

y

B 30°

A x

O

Solution: mg α = −0.5k rad/s2

y

rG/O = 3i + 3.75j m G

F = −F cos30 ◦ i + F sin30 ◦ j

B

rG/O

rB/O = 4.5i + 3.5j m

  +

Fx :

Fy :



Ox − F cos30

◦

=0

(1)

◦

Oy + F sin30 − mg = 0

M0 :

rB/o

Ox

x

(2)

rG/O × (−mg j) + rB/O × F

=

Oy

I0 α

where r G/O × (−mg j) = −3 mgk ◦

◦

rB/O × F = (4.5 F sin30 + 3 F cos 30 )k (−3 mg + 4.5 Fsin30

◦

◦

+ 3 F cos 30 )k = I0 (−0.5k)

(3)

2

where I 0 = 78000 kg-m

Solving Eqns. 1, 2, and 3, we get Ox = 5.70 kN , Oy = 21.3 kN, F

= 6.55 kN .


1

Problem 18.26 Arm BC has a mass of 12 kg and the moment of inertia about its center of mass is 3 kg-m 2 . Point B is stationary and arm BC has a constant counterclockwise angular velocity of 2 rad/s. At the instant shown, what are the couple and the components of force exerted on arm B C at B ?

y

C 0 30 m m 40°

A

x

B

700 mm

Solution: Since the angular acceleration of arm

BC is zero, the sum of the moments about the fixed point B must be zero. Let MB be the couple exerted by the support at B . Then

MB + rCM/B × mg = MB +

 i  0.3cos40 0

◦

j 0.3sin40 ◦ −117.7

k 0 0

  = 0.

MB = 27.05k (N-m) is the coup le exert ed at B . From Newton’s

second law: Bx = max , By − mg = may where ax , ay are the accelerations of the center of mass. From kinematics: a = α × rCM/O − ω 2 rCM/O = −(22 )(i0.3cos40

◦

◦ + j0.3sin40 )

= −0.919i − 0.771j ( m/s2 ),

where the angular accelerat ion is zero from the problem statemen t. Substitute into Newton’s second law to obtain the reactions at B : Bx = −11.0 N , By = 108.5 N .


1

Problem 18.27 Arm BC has a mass of 12 kg and the moment of inertia about its center of mass is 3 kg-m 2 . At the instant shown, arm AB has a constant clockwise angular velocity of 2 rad/s and arm BC has counterclockwise angular velocit y of 2 rad/s and a clockwise angular acceleration of 4 rad/s 2 . What are the coupl e and the components of force exerted on arm BC at B ? Solution: Because the point

B is accelerating, the equations of angular motion must be written about the center of mass of arm B C . The vector distances from A to B and B to G , respectively, are By

rB/A = rB − rA = 0.7i,

MB

◦

◦

rG/B = 0.3 cos(40 )i + 0.3sin (40 ) j

mg Bx

= 0.2298i + 0.1928j ( m).

The acceleration of point B is 2 2 aB = α × rB/A − ωAB rB/A = −ωAB (0.7i) (m/s2 ).

The acceleration of the center of mass is aG = aB +

aG

αBC

2 × rG/B − ωBC rG/B

 = −2.8i + 

i j 0 0 0.2298 0 .1928

k −4 0

  − 0.9193i − 0.7713j

= −2.948i − 1.691j (m/s2 ).

From Newton’s second law, Bx = maGx = (12)(−2.948) = −35.37 N , By − mg = maGy , By = (12)(−1.691) + (12)(9.81) = 97.43 N

From the equation of angular motion, MG = IαBC . The moment about the center of mass is

MG = MB + rB/G × B =

 i  −0.2298 −35.37

j −0.1928 97 .43

k 0 0

 

= MB k − 29.21k (N-m). 2

Note I = 3 kg-m and

αBC

= −4k ( rad/s2 ), from which

MB = 29.21 + 3(−4) = 17.21 N-m .


1

Problem 18.28 The space shuttle’s attitude control engines exert two forces Ff = 8 kN and Fr = 2 kN. The force vectors and the center of mass G lie in the x -y plane of the inertial reference frame. The mass of the shuttle is 54,000 kg, and its moment of inertia about the axis through the center of mass that is parallel to the z axis is 4 .5 × 106 kg-m2 . Determine the acceleration of the center of mass and the angular acceler ation. (You can ignore the force on the shuttle due to its weight).

y 2m 2m

Ff

Fr G

5° m 18

6° 12m

x

Solution: Newton’s second law is



F = (Ff cos 5◦ − Fr cos 6◦ )i − (Ff sin5 ◦ + Fr sin 6◦ ) j = ma.

Setting Ff = 8000 N, Fr = 2000 N and m = 54,000 kg and solving for a , we obtain a = 0.1108i − 0.0168j ( m/s2 ). The equation of angular motion is



M = (18)(Ff sin5 ◦ ) − (2)(Ff cos 5◦ ) ◦ ◦ − (12)(Fr sin6 ) + (2)(Fr cos 6 ) = I α

where I = 4.5 × 106 kg-m2 . Solving for α the counterclockwise angular acceleration is α = −0.000427 rad/s2 .


1

Problem 18.29 In Problem 18.28, suppose that Ff = 4 kN and you want the shuttle’s angular acceleration to be zero. Determine the necessary force Fr and the resulting acceleration of the center of mass. Solution: The total moment about the cent er of mass must equal zero:



M = (18)(Ff sin5 ◦ ) − (2)(Ff cos 5◦ ) ◦ ◦ − (12)(Fr sin6 ) + (2)(Fr cos 6 ) = 0

Setting Ff = 4000 N and solving Fr = 2306 N. From Newton’s second law



◦ ◦ F = (Ff cos 5 − Fr cos 6 )i ◦

◦

− (Ff sin5 + Fr sin 6 ) j = 54,000a,

we obtain a = 0.0313i − 0.0109j (m/s2 ).


1

Problem 18.30 Points B and C lie in the x -y plane. The y axis is vertical. The center of mass of the 18B to kg arm BC is at the midpoint of the line from C , and the moment of inertia of the arm about the axis through the center of mass that is parallel to the z axis is 1 .5 kg-m 2 . At the instant shown, the angular velocity and angular acceleration vectors of arm AB are ωAB = 0.6k (rad/s) and αAB = −0.3k (rad/s2 ). The angular velocity and angular acceleration vectors of arm B C are ωBC = 0.4k (rad/s) and αBC = 2k (rad/s)2 . Determine the force and couple exerted on arm B C at B .

y

760 m

A

15°

50

0 90

m

m x

B

z

Solution: The acceleration of point

C

m

B is aB = aA + αAB ×

2 rA/B − ωAB rA/B or

mg

aB =

 i  0  0.76cos15 ◦

j 0 ◦ −0.76sin15

k −0.3 0

◦

  

By

50

◦

− (0.6)2 (0.76cos15 i − 0.76sin15 j)

°

Bx MB

= −0.323i − 0.149j ( m/s2 )

The accelerat ion of the center of mass G of arm B C is aG = aB +

αBC

2 rG/B × rG/B − ωBC

 i  +  0 0.45cos50 ◦

j

0 0.45sin50 ◦

aB = −0.323i − 0.149j

  2  0 k

◦ ◦ − (0.4)2 (0.45cos50 i + 0.45sin50 j),

or a G = −1.059i + 0.374j ( m/s2 ). The free body diagram of arm B C is: Newton’s second law is



F = Bx i + (By − mg) j = maG :





Bx i + By − (18)(9.81) j = 18(−1.059i + 0.374j).

Solving, we obtain B x = −19.1 N, B y = 183.3 N. The equation of angular motion is

 or

MG = IBC αBC :

(0.45sin50 ◦ )Bx − (0.45cos50 ◦ )By + MB = (1.5)(2)

Solving for MB , we obtain M B = 62.6 N-m.


1

Problem 18.31 Points B and C lie in the x -y plane. The y axis is vertical. The center of mass of the 18kg arm BC is at the midpoint of the line from B to C , and the moment of inertia of the arm about the axis through the center of mass that is parallel to the z axis is 1 .5 kg-m 2 . At the instant show n, the angular velocity and angular acceleration vectors of arm AB are ωAB = 0.6k (rad/s) and αAB = −0.3k (rad/s2 ). The angular velocity vectors of arm BC is ωBC = 0.4k (rad/s). If you want to program the robot so that the angular acceleration of arm BC is zero at this instant, what couple must be exerted on arm BC at B ? Solution: From the solution of Problem 18.30, the acceleration of point B is aB = −0.323i − 0.149j ( m/s2 ). If α BC = 0, the acceleration of the center of mass G of arm B C is 2 rG/B = −0.323i − 0.149j aG = aB − ωBC ◦ ◦ − (0.4)2 (0.45cos50 i + 0.45sin50 j)

= −0.370i − 0.205j (m/s2 ).

From the free body diagram of arm BC in the solution of Problem 18.30. Newton’s second law is



F = Bx i + (By − mg) j = maG :

Bx i + [By − (18)(9.81)] j = 18(−0.370i − 0.205j).

Solving, we obtain Bx = −6.65 N, By = 172.90 N. The equation of angular motion is



MG = IBC αBC = 0:

(0.45sin50 ◦ )Bx − (0.45cos50 ◦ )By + MB = 0.

Solving for MB , we obtain M B = 52.3 N-m.


1

Problem 18.32 The 9000-kg airplane has just landed. At the instant shown, its angular velocity is zero. Its landing gear are rolling and contact the runway at x = 10 m. The friction force on the wheels is negligible. The coordinates of the airplane’s center of mass are x = 10.50, y = 3.00 m. The total aerodynamic force is −26.8i + 30.4j ( kN), and it effectively acts at the center of pressure located at x = 10.75, y = 3.2 m. The thrust T = 4.40 kN exerts no moment about the center of mass. The moment of inertia of the airplane about its center of mass is 75,000 kg-m 2 . Determine the airplane’s angular acceleration.

y T

15

x

Strategy: Draw a free-bod y diagram of the airplan e, including the normal force exerted on the landing gear. To relate the acceleration of the center of mass to the angular acceleration, use the fact that the acceleration of

the airplane (treated as a rigid body), is horizontal at the point where the wheels contact the runway. Solution: The free body diagram is as shown. C is the center of pressure and A is the aerodynamic force. Newton’s second law is



y

A G

F = T + A + N − mg j = ma:

mg j P

◦

T C

◦

T cos15 i + T sin15 j + Ax i + Ay j + N j − mg j = m(ax i + ay j).

15

°

x

N

Equating i and j components, 4400 cos 15◦ − 26,800 = 9000ax , 4400sin 15◦ + 30,400 + N − (9000)(9.81) = 9000ay

(1).

The moment of A about the center of mass is

  i 0.25  −26,800

rC/G × A = 



j k 0.20 0  = 12,960k ( N-m). 30 ,400 0 

The equation of angular motion is



M = 12,960 − (0.5)N = 75,000α.

(2)

To relate ax , ay and α we express the acceleration of the point of   j k   i contact P as a p i = aG + α × rp/G = ax i + ay j +  0 0 α .  −0.5 −3 0  Equating the j components, 0 = ay − 0.5α

(3)

Solving Equations (1)–(3), the angular acceleration is α = −0.200 rad/s2


1

Problem 18.33 The radius of the 2-kg disk is R = 80 mm. It is released from rest on the rough inclined surface.

(a) (b)

R

How long doe s it tak e the disk to roll thr ough one revolution? What minimum coefficient of static friction between the disk and the surface is necessary for the disk to roll instead of slipping when it is released?

30

◦ Solution: Given: R = 80 mm , g = 9.81 m/s2 , θ = 30 , no slip

We have the dynamic, kinematic, and no slip equations



F : F − mg sin θ = −ma



F : N − mg cos θ = 0



mg

1 MG : −F R = − mR 2 α 2

F

a = Rα N

30°

F = µN Solving we find

α=

(a)

g 3R

α=

, a=

g 3

9.81 m/s2 3(0.08 m )

, N = 0.867 mg, F = 0.167 mg, µ = 0.192

= 40.9 rad/s2 , ω = (40.9 rad/s2 )t, θ = (20.4 rad/s2 )t 2

For one revolution θ = 2π rad = (20.4 rad/s2 )t 2 ⇒ t = 0.554 s (b)

µ = 0.192

c 2005 Pearson Educat ion, Inc., Upper Saddle River, NJ. All rights reserve d. This material is protected under all copyright laws as they  currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Problem 18.34 A thin ring and a circular disk, each of mass m and radius R , are released from rest on an inclined surface and allowed to roll a distance D . Determine the ratio of the times required to traverse the distance D in the two cases.

Solution: Choose a coordinate system with the x axis parallel to

= R×f=

 −

i 0 f

j

−R 0

k 0 0

 =−

R

D

D

y

the incl ined surface and the origin at the center of the disk at the instant of release. The moment about the center of mass is

M

R

W f

Rf k.

θ

N α

From the equation of angular motion, M

=I

, from which

−Rf = Iα,f = −RI α .

The time required to travel a distance D after being released from rest is

 = 2D

2D(I

+ R 2 m) .

From Newton’s second law and the free body diagram:

t



The moment of inertia for a thin ring of radius R and mass m about mR 2 . The time to travel a distance D is the polar axis is I ring

Fx

= −f + mg sin θ = max ,

=

=

where ax is the acceleration of the center of mass. Substitute the expression for the force f :

tring Iα R

+ mg sin θ = max .

The relation between α and a x is found from kinematics: the acceleration of the point of contact P with the inclined surface is zero, from which 0

= aCM + × rP /C − ω2 rP C = aCM + × (−R) j − ω2 (−R j), α

α

and

aCM

 = 2

D g sin θ

.

The moment of inertia of a disk of radius 1 mR 2 . polar axis is I disk 2

R and mass m about the

=

The time to travel a distance D is

tdisk

=



3D

g sin θ

.

The ratio of the times is

= × R j − ω2 R j α

 =

i 0 0

j 0 R

k α 0

From the constraint on the motion, aCM or

α

R 2 mg sin θ

ax

 −

ω2 R j

= −αR i − ω2 R j.

tring tdisk

= √2 = 3



4 3

= ax i, from which ax = −Rα ,

= − aRx .

Substitute and solve:

ax

=



mg sin θ . I m R2

+





1

Problem 18.35

The stepped disk weighs 40 lb and it s moment of inertia is I = 0.2 slug-ft2 . If the di sk is released from rest, how long does it take its center to fall 3 ft? (Assume that the string remains vertical.)

4 in 8 in

Solution: The mom ent abo ut the cente r of mass is M = −RT . From the equation of angular motion: −RT = I α, from which T = Iα . From the free body diagram and Newton’s second law: Fy = − R T − W = may , where ay is the acceleration of the center of mass. From kinematics: a y = −Rα . Substitute and solve:

T



ay =



I

W

R2

+m



. W

The time required to fall a distance D is

t=

  2D

ay

=

2D(I + R 2 m)

R2W

.

4 W = 0.3333 ft, W = 40 lb, m = = 1.24 slug, 12 g I = 0.2 slug-ft , t = 0.676 s For D = 3 ft, R = 2


1

Problem 18.36 The radius of the pulley is R= 100 mm and its moment of inertia is I = 0.1 kg-m 2 . The mass m = 5 kg. The spring constant k = 135 N/m. The system is released from rest with the spring unstretched. Determine how fast the mass is moving when it has fallen a distance x = 0.5 m. Strategy: Draw individual free-body diagrams of the mass and pulley, and use them to determine the acceleration a of the mass as a function of the distance x it has fallen. Then use the chain rule: a = dv/dt = (dv/dx)(dx/dt) = (dv/dx)v .

R

k

m x

Solution: T1 = +kx Oy



m0 :

− RT1 + RT2 = I0 α Ox

ax = Rα T1



Fy :

T2

− T2 + mg = max

Combining equations

−T1 + T2 =

+

   I0 R

ax R

+

x

x

−T2 + mg = may

Adding the two equations, we get

−T1 + mg =



m+

I0 R2



T2

ax +

x

where T 1 = +kx , a x = v dv dx mg



m+

 

I0 R2



m+

m+

I0 R2

v

dv = −kx + mg dx

I0 R2



 

v

v du =

=

v2 = 2

0



−k



+0.5

(−kx + mg)dx 0

x2 + mgx 2

v = 1.01 m/s

 

+0.5 0


1

The radius of the pulley is R= 100 mm and its moment of inertia is I = 0.1 kg-m 2 . The mass m = 5 kg. The spring constant is k = 135 N/m. The system is released from rest with the spring unstretched. What maximum velocity does the mass attain as it falls? Problem 18.37

Solution: From the solution to Problem 18.36,

1 2



m+

I0 R2



R

v 2 = mgx − kx 2 /2

Let c = (m + I0 /R2 ) and multiply the expression by 2.

cv 2 = 2mgx − kx 2 c 2v

dv

= 2mg − 2kx

dx Setting

k

m x

dv mg = 0, x = dx k

Evaluating v at this value of x yields

2 cvmax =2

2 vmax =

m2 g 2 m2 g2 k − k k2

m2 g2 , or v max = 1.09 m/s ck


1

Problem 18.38

The mass of the di sk is 45 kg and its radius is R = 0.3 m. The spring constant is k = 60 N/m. The disk is rolled to the left until the spring is compressed 0.5 m and released from rest. (a)

R

k

If you assume that the disk rolls, what is its angular acceleration at the instant it is released? What is the minimum coefficient of static friction for which the disk will not slip when it is released?

(b)

Solution: +

x0 = −0.5

mg

k = 600 N/m

Fs

O

m = 45 kg R = 0.3 m I0 =

1 mR2 2

f

= 2.025 N-m2 , Fs = kx



Fx :

− Fs − f = ma0x



Fy :

N − mg = 0



M0 :

x

N

− f R = I0 α

Rolling implies a 0x = −Rα We have, at x = −0.5 m − kx − f = ma0x

N − mg = 0

− Rf = I0 α

a0x = −Rα Four eqns, four unknowns (a0x , α,N,f) (a)

Solving f = 100 N, N = 441.5 N

α = −14.81 rad/s2 (clockwise) a0x = 4.44 m/s2 (b)

for impending slip,

f = µs N µs = f/N = 100/441.5 µs = 0.227


1

Problem 18.39

The disk weighs 12 lb and its radiu s is R = 6 in. The spring constant is k = 3 lb/ft. The disk is released from rest with the spring unstretched. Determine the magnitude of the velocity of the center of the disk when it has moved 2 ft from its initial position if (a) the inclined surface is smooth (friction is negligible); (b) the disk rolls on the surface.

k

R

30

Solution: We draw the FBD after the disk has moved an arbitrary distance x down the ramp.

kx

Given:

W = 12 lb , g = 32.2 ft/s2 , m = W/g

θ

mg ◦

k = 3 lb/ft, R = 6 in. , θ = 30

F F

For both cases we have 

N



: N − mg cos θ = 0

M

G

(a)

F

: F + kx − mg sin θ = −ma

1 : −F R = − mR2 α 2

The ramp is smooth ⇒ F = 0. Solving we find

a=v



dv dx

= (16.1 ft/s2 ) − (8.05 s −2 )x

v

vd v = 0



2 ft

[(16.1 ft/s2 ) − (8.05 s −2 )x ]d x

0

v2 = (16.1 ft/s2 )(2 ft) − (4.025 s−2 )(2 ft)2 = 16.1 ft2 /s 2 2 v = 5.67 ft/s (b)

The disk rolls w/o slipping ⇒ a = Rα . Solving we find

a=v



dv dx

= (10.73 ft/s2 ) − (5.37 s −2 )x

v

vd v = 0



2 ft

[(10.73 ft/s2 ) − (5.37 s −2 )x ]d x

0

v2 = (10.73 ft/s2 )(2 ft) − (2.68 s −2 )(2 ft)2 = 10.73 ft2 /s 2 2 v = 4.63 ft/s

c 2005 Pearson Educat ion, Inc., Upper Saddle River, NJ. All rights reserve d. This material is protected under all copyright laws as they  currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Problem 18.40 A 42-lb sphere with radius R = 4 in. is placed on a horizontal surface with initial angular velocity ω0 = 40 rad/s. The coefficient of kinetic friction between the sphere and the surface is µk = 0.06. What maximum velocity will the center of the sphere attain, and how long does it take to reach that velocity?

ω

0

Strategy: The friction force exerte d on the spinning sphere by the surface will cause the sphere to accelerate to the right. The friction force will also cause the sphere’s angular velocity to decrease. The center of the sphere will accelerate until the sphere is rolling on the surface instead of slipping relative to it. Use the relation between the velocity of the center and the angular velocity of the sphere when it is rolling to determine when the sphere begins rolling. Solution: Given W = 42 lb , g = 32.2 ft/s2 , m = W/g, R = 4/12 ft, µk = 0.06 We have



Fx : µk N = ma



Fy : N − mg = 0



W

W

MG : µk NR =

2 5

µ RW

2

mR a

Solving we find

α=

5µk g 2R

= 14.49 rad/s2 , a = µk g = 1.932 ft/s2

From kinematics we learn that

W

α = 14.49 rad/s2 , ω = (14.49 rad/s2 )t − (40 rad/s) a = 1.932 ft/s2 , v = (1.932 ft/s2 )t when we reach a steady motion we have

µk N

N

v = −Rω ⇒ (1.932 ft/s2 )t = −(0.33 ft)[(14.49 rad/s2 )t − (40 rad/s)] Solving for the time we find

t = 1.97 s ⇒ v = 3.81 ft/s


1

Problem 18.41

A soccer pla yer kicks the ball to a teammate 8 m away. The ball leaves the player’s foot moving parallel to the ground at 6 m/s with no angular velocity. The coefficient of kinetic friction between the ball and the grass is µk = 0.32. How long does it take the ball to reach his teammate? The radius of the ball is 112 mm and its mass is 0.4 kg. Estimate the ball’s moment of inertia by using the equation for a thin spherical shell: I = 23 mR 2 .

Solution: Given µ = 0.32, r = 0.112 m, g = 9.81 m/s2 , v0 = 6 m/s

The motion occurs in two phases. (a)

Slipping.



Fx : −µN = ma



Fy : N − mg = 0



MG : −µN R =

mg

µΝ

N

2 3

mR 2 α

Solving we find

a = −µg ⇒ v = v0 − µgt, s = v0 t −

α=−

3µg 2R

⇒ ω=−

3µg 2R

1 2

µgt 2

t

When it stops slipping we have 3 2v0 v = −Rω ⇒ v0 − µgt = 2 µgt ⇒ t = 5µg = 0.765 s

v = 3.6 m/s, s = 3.67 m (b)

Rolling— Steady motion

a = 0, v = 3.6 m/s, s = (3.6 m/s)(t − 0.765 s) + 3.67 m When it reaches the teammate we have 8 m = (3.6 m/s)(t − 0.765 s) + 3.67 m ⇒ t = 1.97 s


1

Problem 18.42 The 100-kg cylindrical disk is at rest when the force F is applied to a cord wrapped around it. The static and kinetic coefficients of friction between the disk and the surface equal 0.2. Determine the angular acceleration of the disk if (a) F = 500 N and (b) F = 1000 N.

F 300 mm

Strategy: First solve the problem by assuming that the disk does not slip, but rolls on the surface. Determine the friction force, and find out whether it exceeds the product of the coefficient of friction and the normal force. If it does, you must rework the problem assuming that the disk slips. Solution: Choose a coordinate system with the srcin at the center of the disk in the at rest position, with the x axis parallel to the plane surface. The moment about the center of mass is M = −RF − Rf ,

F

from which − RF − Rf = I α. From which

f =

−RF − I α

R

= −F −

W

Iα . R

f

From Newton’s second law: F − f = max , where a x is the acceleration of the center of mass. Assume that the disk rolls. At the point of contact a P = 0; from which 0 = aG + α × rP /G − ω2 rP /G.

(b)

N For F = 1000 N the acceleration is

ax = aG = ax i = α × R j − ω2 R j

4F 3m

=

4000 300

= 13.33 m/s2 .

The friction force is =

 

i 0 0

j 0 R

k α 0

 

− ω2 R j = −Rα i − ω2 R j,

f = F − max = 1000 − 1333.3 = −333.3 N.

from which ay = 0 and a x = −Rα. Substitute for f and solve:

ax =

2F



m+

I R2



.

The drum slips . The moment equation for slip is Rµk gm = I α, from which

α=

For a disk, the moment of inertia about the polar axis is

I =

−RF + Rµk gm

I

=−

−RF +

2F 2µk g + = −53.6 rad/s2 . mR R

1 mR2 , 2

from which

ax = (a)

4F 3m

=

2000 300

= 6.67 m/s 2 .

For F = 500 N, the friction force is

f = F − max = −

500 F =− = −167 N . 3 3

Note: −µk W = −0.2 mg = −196.2 N, the disk does not slip . The angular velocity is

α=−

ax 6.67 =− = −22.22 rad/s2 . 0.3 R


1

Problem 18.43

The ring gear is fixed. The ma ss and moment of inertia of the sun gear are mS = 320 kg and 2 IS = 40 kg-m . The mass and moment of inertia of each planet gear are mP = 38 kg and IP = 0.60 kg-m2 . If a couple M = 200 N-m is applied to the sun gear, what is the latter’s angular acceleration?

Ring gear

0.18 m.

0.86 m

M

0.50 m

Planet gears (3)

Sun gear Solution: MS

F

= 200 N-m

Sun Gear:



Planet Gears:



er

M0 :

act

r

Gr

G

− F r = IP αP

R

F

IP

O

et

Ft :

From kinematics

− 3RF = IS αS C

Mc :



MS

F

+ G = mP act

3 small disks

Is

Ms F F

= −rα P

2αP rP = −RαS We have 5 eqns in 5 unknowns. Solving, clockwise)

αS

= 3.95 rad/s2 (counter-


1

Problem 18.44

In Problem 18.43, what is the magnitude of the tange ntial force exerted on the sun gear by each planet gear at their point of contact when the 200 N-m couple is applied to the sun gear? Solution: See the soluti on to Problem 18 .43. Solving the 5 eqns in 5 unknowns yields αS = 3.95 rad/s2 , G = 9.63 N, aGt = 0.988 m/s2 , αP = −5.49 rad/s2 , F = 27.9 N. F is the required value.


1

Problem 18.45 The 18-kg ladder is released from rest in the position shown. Model it as a slender bar and neglect friction. At the instant of release, determine (a) the angular acceleration of the ladder and (b) the normal force exerted on the ladder by the floor. 30° 4m

Solution: The vect or location of the cent er of mass is rG =

(L/2) sin30 ◦ i + (L/2) cos30 ◦ j = 1i + 1.732j ( m). Denote the normal forces at the top and bottom of the ladder by P and N . The vector locations of A and B are rA = L sin30 ◦ i = 2i ( m), rB = L cos30 ◦ j = 3.46j ( m). The vecto rs rA/G = rA − rG = 1i − 1.732j ( m), rB/G = rB − rG = −1i + 1.732j (m). The moment about the center of mass is

M = rB/G × P + rA/G × N,

i M =  −1

j 1.732 0

P

B

mg

 i k 0 +1 0

P

0

j −1.732 N

 k 0

A N

0

= (−1.732P + N )k ( N-m).

From the equation of angular motion: (1) − 1.732 P + N = I α . From Newton’s second law: (2) P = max , (3) N − mg = may , where ax , ay are the accelerations of the center of mass. From kinematics: aG = aA + α × rG/A − ω 2 rG/A . The angular velocity is zero since the system was released from rest,

aG = aA i +

 

i 0 −1

j 0 1.732

k α 0

 =a

Ai

− 1.732α i − α j

= (aA − 1.732α) i − α j ( m/s2 ),

from which a y = −α . Similarly,

aG = aB +

α

× rG/B , aG = aB +

i 0 1

j 0 −1.732

k α 0

 

= aB j + 1.732α i + α j,

from which a x = 1.732α . Substitute into (1), (2) and (3) to obtain three equations in three unknowns: −1.732P + N = I α , P = m(1.732)α , N − mg = −mα . Solve: (a) α = 1.84 rad/s2 , P = 57.3 N, (b) N = 143.47 N


1

Problem 18.46 The 18-kg ladder is released from rest in the position shown. Model it as a slender bar and neglect friction. Determine its angular acceleration at the instant of release. 30 4m

20

Solution: Given m = 18 kg ,

L = 4 m, g = 9.81 m/s2 , ω = 0 NA

First find the kinematic constraints. We have aA = aG + α × rA/G

= ax i + ay j + α k ×

=



ax − α

L

cos30

2



−

◦

L

2

sin30

◦

  i+

 

i + ay − α

L

2

L

2

sin30

cos30

◦



◦

30°

 j

j mg

aB = aG + α × rB/G

= ax i + ay j + α k ×

=



ax + α

L

cos30

2



◦

L

sin30

2

◦

 

i+ −

 

i + ay + α

L

2

L

2

sin30

cos30

◦



◦

 j

NB

j

20°

The constraints are aA · i = ax − α

L

2

◦

cos30 = 0

◦

◦

aB · (sin20 i + cos20 j)

=



ax + α

L

cos30

2



◦



◦

sin20 + ay + α

L

2

sin30

◦



cos20

◦

The dynamic equations:

  

Fx : N A + NB sin20

◦

= max

◦

Fy : N B cos20 − mg = may

MG : −NA



L

2

cos30

+ NB sin20

◦



◦

L

2



+ NB cos20

cos30

◦



=

◦

1 12



L

2

sin30

◦



mL2 α

Solving five equations in five unknowns we have α = 2.35 rad/s2

CC W

Also ax = 4.07 ft/s2 , ay = −5.31 ft/s2 , NA = 43.7 N , NB = 86.2 N


1

Problem 18.47 The 4-kg slender bar is released from rest on the rough surface in the position shown. Determine the minimum value of the coefficient of static friction for which the bar will not slip relative to the floor when it is released. 1m

60°

Solution: Given m = 4 kg,

L = 1 m, g = 9.81 m/s2

Kinematics: aB = aG + α × rB/G



= ax i + ay j + α k ×

=



ax + α

L

2

sin60

◦

L

2

◦

cos60 i −

 

i + ay + α

L

2

L

2

◦

G



sin60 j

cos60

◦



j mg

The contact with the floor requires aB · i = ax + α

aB · j = ay + α

L

2 L

2

◦

sin60 = 0

B µNB

◦

cos60 = 0

The dynamic equations

  

60°

NB

Fx : −µNB = max

Fy : N B − mg = may

MB : mg

L

2

Solving we find

◦

cos60 =

1 3

mL2 α

µ = 0.400

We also have ax = −3.19 m/s2 , ay = −1.84 m/s2 , α = 7.36 rad/s2 , NB = 31.9 N


1

Problem 18.48 The masses of the bar and disk are 14 kg and 9 kg, respectively. The system is released from rest with the bar horizontal. Determine the bar’s angular acceleration at that instant if

A O 1.2 m

(a) (b)

the bar and disk are welded together at A, the bar and disk ar e conn ected by a smooth pin at A.

0.3 m

Strategy: In Part (b), draw individ ual free-body diagrams of the bar and disk. Solution: (a)

L = 1.2 m

R = 0.3 m

mB = 14 kg

Ox

mD = 9 kg

O

C G

A mog

Oy

mBg

O is a fixed point For the bar IG =

1 12

1

mB L 2 =

IOB = IG + mB

12

(14)(1.2)2 = 1.68 N-m 2

2

  L 2

IOB = 6.72 N-m 2 For the disk:

IA =

1 m R2 2 D

=

1 (9)(0.3)2 2

= 0.405 N-m2

IOD = IA + m0 L2 = 13.37 N-m2 The total moment of inertia of the welded disk and bar about O is

IT = IOB + IOD = 20.09 N-m 2

  

Fx :

Ox = O = maGx

Fy :

Oy − mB g − mD g = (mB + mD )aGy

M0 :

−

  L 2

mB g − LmD g = IT α

We can solve the last equation for α without finding the location and acceleration of the center of mass, G . Solving,

α = −9.38 rad/s2 (b)

(clockwise)

In this case, only the moment of inertia changes. Since the disk is on a smooth pin, it does not rotate. It acts only as a point mass at a distance L from point O .   In this case, I OD = mD L2 and I T = IOB + IOD = 19.68 N-m2 We now have



M0 :

−

  L 2

mB g − LmD g = IT α 

Solving α  = −9.57 rad/s2

(clockwise)


1

Problem 18.49

The 5-lb horizo ntal bar is connect ed to the 10-lb disk by a smooth pin at A. The system is released from rest in the position shown. What are the angular accelerations of the bar and disk at that instant?

A O

3 ft

1 ft

Solution: Given

Ay

g = 32.2 ft/s2 , Wbar = 5 lb, Wdisk = 10 lb , mbar =

Wbar g

, mdisk =

Ax

Ox

Wdisk g

Oy Wbar

L = 3 ft, R = 1 ft

Ay

Wdisk

The FBDs The dynamic equations



MO : −mbar g



MGdisk : −Ay R =



Fy : A y − mdisk g = mdisk aydisk

L 2

− Ay L =

1 2

1 3

mbar L2 αbar

mdisk R 2 αdisk

Kinematic constraint

αbar L = aydisk − αdisk R Solving we find

αdisk = 3.58 rad/s2 , αbar = −12.5 rad/s2 , aydisk = −34.0 m/s2 , Ay = −0.556 N Thus

αdisk = 3.58 rad/s2

CCW, α

bar

= 12.5 rad/s2

CW


1

Problem 18.50 The 0.1-kg slender bar and 0.2-kg cylindrical disk are released from rest with the bar horizontal. The disk rolls on the curved surface. What is the bar’s angular acceleration at that instant it is released?

40 mm

120 mm

Solution: The mom ent abo ut the cente r of mass of the dis k is M = f R, from the equation of angular motion, Rf = Id αd . From Newton’s second law: f − By − Wd = md ady . Since the disk rolls, the kinematic condition is ady = −Rαd . Combine the expressions and rearrange: f = I αd /R, I αd /R − By − Wd = md ady , from which By + Wd = (Rmd + Id /R) αd . The moment about the center of mass of the bar is

 

Ay +

 

 

By = Ib αb .

L 2

Mb = −

L 2

Ax

Ay

By

By Bx

f N

Bx Wb Wd

By ,

from which −

  L 2

L 2

Ay +

From Newton’s second law Ay − Wb + By = mb aby , where aby is the acceleration of the center of mass of the bar. The kinematic condition for the bar is aCM = αb ×

     L 2

i

L 2

=

αb j,

from which

aby =

  L 2

αb .

Similarly, a D = aCM +

αb

× ((L/2)i), from which a dy = Lαb .

From which: αd = −Lαb /R. Substitute to obtain three equations in three unknowns:

By + W d =

−

  L 2



Rmd +

Ay +

  L 2

Ay − W b + By = m b

Id R

  −

L R

αb ,

By = Ib αb ,

  L 2

αb .

Substitute known numerical values: L = 0.12 m, R = 0.04 m, m b = 0.1 kg, Wb = mb g = 0.981 N, md = 0.2 kg, Wd = md g = 1.962 N, Ib = (1/12)mb (L2 ) = 1.2 × 10−4 kg-m2 , Id = (1/2)md R 2 = 1.6 × 10−4 kg-m2 . Solve:

αb = −61.3 rad/s2 , Ay = 0.368 N , By = 0.245 N .


1

The mass of the suspended object A is 8 kg. The mass of the pulley is 5 kg, and its moment of inertia is 0 .036 kg-m 2 . If the force T = 70 N, what is the magnitude of the acceleration of A ? Problem 18.51

T

120 mm

A

Solution: Given

T2 T

mA = 8 kg, mB = 5 kg, IB = 0.036 kg-m2 R = 0.12 m , g = 9.81 m/s2 , T = 70 N

m Bg




FyB : T2 + T − mB g − By = mB aBy



FyA : By − mA g = mA aAy



MB : −T2 R + T R = IB αB

By

By

Kinematic constraints

aBy = aAy, aBy = RαB Solving we find

aAy = 0.805 m/s2

mAg

We also have

aBy = 0.805 m/s2 , αB = 6.70 rad/s, T2 = 68.0 N , By = 84.9 N


1

The suspended object A weighs 20 lb. The pulleys are identical, each weighing 10 lb and having moment of inertia 0 .022 slug-ft2 . If the force T = 15 lb, what is the magnitude of the acceleration of A? Problem 18.52

T

4 in

4 in

A

Solution: Given

Wdisk

T2

g = 32.2 ft/s2 , WA = 20 lb , Wdisk = 10 lb , I = 0.022 slug-ft2 mA =

WA g

, mdisk =

Wdisk g

, R=

4 12

T

ft, T = 15 lb




Fy 1 : T2 + T − T1 − mdisk g = mdisk a1



Fy 2 : T4 + T1 − T3 − mdisk g = mdisk a2



Fy 3 : T3 − mA g = mA aA

T4

Wdisk T1

 M1 : T R − T2 R = I α1 

T3

M2 : T1 R − T4 R = I α2

The kinematic constraints

WA

a1 = Rα1 , a2 = Rα2 , a1 = 2Rα2 , aA = a2 Solving we find

aA = 3.16 ft/s2

We also have

a1 = 6.32 ft/s2 , a2 = 3.16 ft.s2 , α1 = 19.0 rad/s2 , α2 = 9.48 rad/s2 T1 = 16.8 lb, T2 = 13.7 lb, T3 = 22.0 lb, T4 = 16.2 lb


1

Problem 18.53 The 2-kg slender bar and 5-kg block are released from rest in the position shown. If friction is negligible, what is the block’s acceleration at that instant?

Solution:

L = 1 m,

m = 2 kg

Assume directions for B x , B y , IG =

    

55°

M = 5 kg

1 12

mB L 2

Fx :

Bx = maGx

(1)

Fy :

By − mg = mB aGy

(2)

L

MG :



cos θ

2

L

By +





Fx :

− Bx = Ma 0x

Fy :

N − By − Mg = 0

sin θ

2

1m

Bx = IG α



y L cos θ 2

(3)

m (4)

G (5)

θ

(L / 2) sin θ

mg

From kinematics, ω = 0 (initially) O

a0 = aG + α k × r0/G

where

r0/G =

L

2

cos θ i −

L

2

x

Bx By

sin θ j

From the diagram a 0 = a0x i



a0x 0

= aGx + (αL/2) sin θ = aGy + (αL/2) cos θ

(6) (7)

We know θ = 55◦ , IG = 0.167 kg-m2 , L = 1 m, m = 2 kg, M = 5 kg. We have 7 eqns in 7 unknowns

Mg

M Bx

By O

(aGx , aGy , a0x ,α,B x , By ,N),

Solving, we get Bx = −5.77 N ,

(opposite the assumed direction)

N

By = 13.97 N , aGx = −2.88 m/s2 , α = 9.86 rad/s2 ,

aGy = −2.83 m/s2 N = 63.0 N

2

a0x = 1.15 m/s . (to the right)


1

Problem 18.54

The 2-kg slender bar and 5-kg block are released from rest in the position shown. What minimum coefficient of static friction between the block and the horizontal surface would be necessary for the block not to move when the system is released? Solution: This solution is very similar to that of Problem 18.53. We add a friction force f = µs N and set a 0x = 0. L=1m

m = 2 kg

L θ

M = 5 kg

IG =

1 12

G mg

mL2 = 0.167 kg-m2

Fx :

Bx = maGx

(1)

 

Fy :

By − mg = maGy

(2)



MG :

Bx By



L

2

cos θ



By +



L

2

sin θ



Bx = IG α

(3)

Mg

(These are the same as in Problem 18.53) Note: In Prob 18.53, B x = −5.77 N (it was in the opposite direction to that assumed). This resulted in a 0x to the right. Thus, friction must be to the left

By

M

 

Fx :

− Bx − µs N = ma0x = 0

(4)

Fy :

N − By − Mg = 0

(5)

Bx

µs N

From kinematics, a0

= aG + α × r0/G = 0

N

O = aGx + (αL/2) sin θ

(6)

O = aGy + (αL/2) cos θ

(7)

Solving 7 eqns in 7 unknowns, we get Bx = −6.91 N,

By = 14.78 N, 2

aGx = −3.46 m/s , N = 63.8 N,

aGy = −2.42 m/s2

α = 8.44 rad/s2

µs = 0.108


1

Problem 18.55 The 0.4-kg slender bar and 1-kg dis k are released from rest in the position shown. If the disk rolls, what is the bar’s angular acceleration at the instant of release?

40°

1m

0.25 m

Solution: Choose a coordinate system with the src in at

B and the x axis parallel to the plane surface. From Newton’s second law applied to the disk,

By

By

W Bx

b

N − Wd − By = 0,

Wd

Bx f

Bx − f = md adx .

N

The moment about the mass center of the disk is Md = −Rf , from which −Rf = Id α . From Newton’s second law applied to the bar: −Bx = mb abx , By − Wb = mb aby , where abx , aby are the accelerations of the center of mass of the bar. The vect or location of the center of mass of the bar is rG/B = rG − 0 = −0.5sin40 ◦ i + 0.5cos40 ◦ j

Substitute known numerical values:

mb = 0.4 kg , md = 1 kg , L = 1 m,

= −0.3214i + 0.3830j (m) .

From the equation of angular motion for the bar:

R = 0.25 m, Wb = 3.924 N , Wd = 9.81 N,

−0.3830Bx + 0.3214By = Ib αb .

From the kinematics, the acceleration of the disk is related to the angular acceleration by adx = −Rαd . The acceleration of the mass center of the bar is ab = ad + αb × rCM/B . From the constraint on the

Ib =

motion of the disk, a d = adx i.

Id =

ab = adx i +

 

i 0 −0.3214

j 0 0 .3830

k αb 0

 

  1

12

1 2



mL2 = 0.0333 kg-m2 ,

mR2 = 0.03125 kg-m2 .

Solving, we obtain N = 12.39 N, B y = 2.58 N,

Bx = 1.26 N,

αb = 10.425 rad/s2 ,

= adx i − 0.3830α i − 0.3214α j,

from which

abx = adx − 0.3830α, aby = −0.3214α.

adx = 0.841 m/s2 , f = 0.4203 N .

Substitute to obtain six equations in six unknowns:

N − Wd − By = 0, −0.3830Bx + 0.3214By = Ib αb , −Bx = mb (adx − 0.3830αb ),

By − Wb = mb (−0.3214αb ), Rf =

Id adx , R

Bx − f = md adx .


1

Problem 18.56 The masses of the slender bar and the crate are 9 kg and 36 kg, respectively. The crate rests on a smooth horizontal surface. If the system is stationary at the instant shown and a counterclockwise couple M 300 N-m is applied to the bar, what is the resulting acceleration of the crate?

=

2m

Solution: L IA

= =

√

M

5m

= 13 (9)5

IA

= 15 kg-m2

T

T cos45 i

rB/A

    

Fx :

Fy :

T sin45 j (N)

Ax

= −88.3 N,

Ay

T

= 115.3 N,

N

2

Ax

+ T cos45 ◦ = maG

Ay

− T sin45 ◦ − mg = ma

(1)

x

Fx :

− T cos45 ◦ = Ma

Fy :

N

aCx

= 0.377 m/s , aBY = 0.755 m/s2 , αBC

= −0.377 rad/s2 ,

= −2.26 m/s 2 .

(3)

+ rB/A × T − 0.5 (mg) = IA α

B Mc

(4)

cx

= 271.6 N,

aGY

= 0.755 rad/s2 ,

(2)

Gy

Mc

α

= 173.2 N, 2

= −0.755 m/s , aBX = −1.51 m/s 2 ,

aGX

MA :

45°

where

aGx aGy

aB

Solving, we get

= − = 1i + 2j m

+ T sin45 ◦ − Mg = 0

From kinematics



2m

1m

1 mL2 3

aG

(5)

y

= aA + αk × rG/A

rG/A

= L2 cos θ i + L2 sin θ j

tan θ

= 2,

θ

aA

G

=0

2

T

mg

= 63.4◦

= −αL sin θ/ 2 = +αL cos θ/ 2

(6) (7)

θ

Ax

x 1

= aA + αk × rB/A

Ay

aBx

= −αL sin θ

(8)

aBy

= αL cos θ

(9)

y

Finally, we need the relationship between a B and a C . Cable B C does not stretch aC

= aB + αBC k × rC/B = aCx i

aC

= aB + αBC k × (2i − 2j)

mg

T

45

°

C x

aCx

= aBx + 2αBC

(10)

O

= aBy + 2αBC

(11)

We have 11 eqns in 11 unknowns

Ax , Ay ,T,a

N Cx , aCy ,

α

aCx , aCy , aBx , aBy , αBC



1

Problem 18.57

In Problem 18.56, determine the resulting acceleration of the crate if the coefficient of kinetic friction between the crate and the horizontal surface is µk = 0.2. Solution: The equations are the same as for the solution to Problem 18.56 except for equation (4). The free body diagram for the crate is µk = 0.2

2m The new equation (4) is ◦

− T cos45 + µk N = Ma cx

(4).

M

Solving the resulting set of eleven equations in eleven unknowns, (the same unknowns as in Problem 18.56), we get Ax = −86.4 N ,

m1

Ay = 173.4 N ,

T = 118.6 N ,

N = 269.3 N , 2

T 2

aGx = −0.278 m/s ,

aGy = 0.139 m/s ,

aBx = −0.556 m/s2 ,

aBy = −0.278 m/s2 ,

α = 0.278 rad/s2 ,

m2

mg

45

°

αBc = −0.139 rad/s2 ,

aCx = −0.834 m/s2 (to the left) . kN

µ

N


1

Problem 18.58 Bar AB rotates with a constant angular velocity of 6 rad/s in the counterclockwise direction. The slender bar BCD weighs 10 lb, and the collar that bar BCD is attached to at C weighs 2 lb. The y axis points upward. Neglecting friction, determine the components of the forces exerted on bar BCD by the pins at B and C at the instant shown.

y

D

8 in

C

12 in 6 rad/s

Solution: The velocity at point

vB = ωAB × rB/A =

 

i 0 8

j 0 0

k 6 0

 

ωBC

x

8 in

6 in

4 in

= 48j in/s = 4j (ft/s).

The velocity at point C is

vC = vB +

B

A

B is

2 )i + (5αBC − 10ωBC 2 ) j. aG = (−288 − 10αBC − 5ωBC

× rC/B = vC i +

 

i j 0 0 6 12

k ωBC 0

 

.

vC = 48j − 12ωBC i + 6ωBC j in/s.

From the constraint on the collar at C , the y component of velocity is zero, from which 48 + 6ωBC = 0, ω BC = −8 rad/s. The acceleration at point B is 2 aB = αAB × rB/A − ωAB rB/A

aG = (−288 − (10)(128) − 5(82 ))i + (5(128) − 10(82 )) j = −1888i (in/s2 ) = −157.33i ( ft/s2 ).

From Newton’s second law and the equation of angular motion applied to the free body diagram of bar BCD and the collar C : for the bar BCD, Bx + Cx = mBC D aG , By + Cy = W,

  1

12

2

Cy −

  5

12

  10

By +

12

Bx −

  2

12

Cx = IG αBC ,

= −36(8i) = −288i ( in/s ) = −24i ( ft/s2 ), since

αAB

and for the collar C , − Cx = mC aC , where the units are to be consistent. Solve:

= 0.

The acceleration at point C is Cx = −mC aC =

aC = aB +

αBC

aC = −288i +

2 rC/B , × rC/B − ωBC

 

i j 0 0 6 12

k αBC 0

 

−

2 ωBC (6i

+ 12j),

aC = (−288 − 6(82 ) − 12αBC )i + (6αBC − 12(82 )) j ( in/s2 ).

From the constraint on the collar, a C = aC i. Separate components to obtain the two equations in two unknowns: aC = −672 − 12αBC ,



2 32.17



Bx = mBC D aG − Cx =

By = −(2)IG αBC +

=−

 

 

1

10

6

32.17

By = −125.7 lb



10 6



(184) = 11.44 lb ,

10 32.17

Bx −



(−157.33) − Cx = −60.3 lb

 1 3

( 102 + 202 )

144

+

Cx +





Bx −

5 3

WBC D

6



 1 3

Cx +

10 6

,

Cy = WBC D − By = 10 − By = 135.7 lb

0 = −768 + 6αBC . Solve:

αBC = 128 rad/s2 ,

Cy

aC = −2208 m/s2 = −184 ft/s2 .

Cx

The acceleration of the center of mass of the bar is aG = aB +

αG

aG = −288i +

2 rG/B . × rG/B − ωBC

 

i j 0 0 5 10

k αBC 0

 

By W

Bx

2 (5i + 10j), − ωBC


1

Problem 18.59

The masses of the slender bars AB and BC are 10 kg and 12 kg, respectively. The angular velocities of the bars are zero at the instant shown and the horizontal force F = 150 N. The horizontal surface is smooth. Determine the angular accelerations of the bars.

A

B

0.4 m

C 0.4 m

Solution: Given

Bx

mAB = 10 kg , mBC = 12 kg , g = 9.81 m/s2 LAB = 0.4 m , LBC =

F

0.2 m

Ax Ay



0.42 + 0.22 m, F = 150 N

mAB g

By

By

The FBDs The dynamic equations LAB

1



MA : −mAB g



FBCx : −Bx − F = mBC aBCx



FBCy : −By − mBC g + N = mBC aBCy



MBCG : (B x − F )(0.2 m ) + (By + N )(0.1 m ) =

2

+ By LAB =

3

F

mBC g

mAB LAB 2 αAB

N

1 12

mBC LBC 2 αBC

The kinematic constraints aBCy = αAB LAB + αBC (0.1 m ) aBC x = αBC (0.2 m ) αAB LAB + αBC (0.2 m ) = 0

Solving we find

αAB = 20.6 rad/s2 , αBC = −41.2 rad/s2

αAB = 20.6 rad/s2

CCW, α

BC

= 41.2 rad/s2

CW

We also find aBC x = −8.23 m/s2 , aBCy = 4.12 m/s2 N = 244 N , Bx = −51.2 N , By = 76.5 N ,


1

Problem 18.60 Let the total mo ment of inertia of the car’s two rear wheels and axle be IR , and let the total IF . The moment of inertia of the two front wheels be radius of the tires is R , and the total mass of the car, including the wheels, is m. If the car’s engine exerts a torque (couple) T on the rear wheels and the wheels do not slip, show that the car’s acceleration is

a=

RT R 2 m + IR + IF

.

Strategy: Isolate the wheels and draw three free-body diagrams. Solution: The free body diag rams are as shown: W e shall write

mBg

three equations of motion for each wheel and two equations of motion for the body of the car: We shall sum moments about the axles on each wheel. Rear Wheel :

Fx



Fx = Fx + fR = mR a,



Fy = NR − mR g − Fy = 0,



MRaxle = RfR − T = IR α = IR −

Gx Fy

Gy

Fy



a R

 Fx

Front Wheel :

Gy

mRg

mF g Gx

T



Fx = Gx + fF = mF a,



Fy = NF − mF g − Gy = 0,



MF axle = RfF = IF α = IF −

fR



a R

NR

fF

NF



Car Body :



Fx = −Fx − Gx = mB a,



Fy = Fy + Gy − mB g = 0.

Summing the y equations for all three bodies, we get NR + NF = (mB + mR + mF )g = mg. Summing the equations for all three bodies in the x direction, we get fR + fF = (mB + mR + mF )a = ma. (1) From the moment equations for the wheels, we get fF = −IF a/R2 and fR = −IR a/R2 + T /R. Substituting these into Eq. (1), we get a = RT /(mR2 + IR + IF ) as required.


1

Problem 18.61

The combined mass of the motorcy cle and rider is 160 kg. Each 9-kg wheel has a 330mm radius and a moment of inertia I = 0.8 kg-m 2 . The engine drives the rear wheel by exerting a couple on it. If the rear wheel exerts a 400-N horizontal force on the road and you do not neglect the horizontal force exerted on the road by the front wheel, determine (a) the motorcycle’s acceleration and (b) the normal forces exerted on the road by the rear and front wheels. (The location of the center of mass of the motorcycle not including its wheels, is shown.)

723 mm

A

B

649 mm 1500 mm

Solution: In the free-body diagrams shown, m w = 9 kg and m = 160 − 18 = 142 kg. Let a be the motorcycle’s acceleration to the right and let α be the wheels’ clockwise angular acceleration. Note that

a = 0.33α.

(1)

Front Wheel :



Fx = Bx + fF = mω a,

(2)



Fy = B y + N F − m ω g = 0 ,

(3)



M = −fF (0.33) = Iα .

(4)

Ax

Bx M

Ay

By

mg

Rear Wheel:



Fx = Ax + fR = mω a,

Ay

(5)



Fy = Ay + NR − mω g = 0,

(6)



M = M − fR (0.33) = I α.

(7)

M

m

NR

By

Ax wg fR

Bx m

wg fF

NF

Motorcycle :



Fx = −Ax − Bx = ma,

(8)



Fy = −Ay − By − mg = 0,

(9)



M = −M + (Ax + Bx )(0.723 − 0.33) + By (1.5 − 0.649) − Ay (0.649) = 0.

(10)

Solving Eqs (1)–(10) with f R = 400 N, we obtain

(a) a = 2.39 rad/s2 and

(b) NR = 455 N , NF = 1115 N .


1

Problem 18.62

In Problem 18.61, if the front wheel lifts slightly off the road when the rider accelerates, determine (a) the motorcycle’s acceleration and (b) the torque exerted by the engine on the rear wheel. Solution: See the solution of Problem 18.61. We set N F = 0 and replace Eq. (4) by fF = 0. Then solving Eqs. (1)–(10), we obtain

(a)

a = 9.34 m/s 2 ,

(b)

M = 516 N-m.


1

Problem 18.63 The moment of inertia of the vertica l handle about O is 0.12 slug-ft 2 . The object B weighs 15 lb and rests on a smooth surface. The weight of the bar AB is negligible (which means that you can treat the bar as a two-force member). If the person exerts a 0.2-lb horizontal force on the handle 15 in. above O , what is the resulting angular acceleration of the handle? A

6 in B

O

12 in

Solution: Let α be the clockwise angular acceleration of the hanF

dle. The acceleration of B is: aB = aA + αAB × rB/A :

aB i = (6/12)α i +

  i  0  1

β

15 in.

j 0 −0.5

 k  αAB  0 

C

6 in. O

we see that αAB = 0 and aB = (6/12)α

(1).

15 lb

The free body diagrams of the handle and objectB are as shown. Note that β = arctan(6/12) = 26.6◦ . Newton’s second law for the object B is C cos β = (15/32.2)aB ,

C

β

(2) N

The equation of angular motion for the handle is (15/12)F − (6/12)C cos β = (0.12)α

(3). 2

Solving Equations (1)–(3) with F = 0.2 lb, we obtain α = 1.06 rad/s

A

aA

= (6/12) B

i

aB


1

Problem 18.64 The bars are each 1 m in length and have a mass of 2 kg. They rotate in the horizontal plane . Bar AB rotates with a constant angular velocity of 4 rad/s in the counterclockwise direction. At the instant shown, bar BC is rotating in the counterclockwise direction at 6 rad/s. What is the angular acceleration of bar BC ?

4 rad/s A

B 6 rad/s aBC

C

Solution: Given m = 2 kg,

◦

L = 1 m, θ = 45

By

The FBD The kinematics

Bx

θ

aB = aA + αAB × rB/A − ωAB 2 rB/A G

= 0 + 0 − (4 rad/s)2 (1 m)i = −(16 m/s 2 )i aG = aB +

αBC

× rG/B − ωBC 2 rG/B

= −(16 m/s 2 )i + αBC k × (0.5 m )(cos θ i − sin θ j) − (6 rad/s)2 (0.5 m )(cos θ i − sin θ j) = (−16 m/s 2 + [0.5 m sin θ ]αBC − [18 m/s2 ]cos θ )i + ([0.5 mcos θ ]αBC + [18 m/s2 ] sin θ ) j

Our kinematic constraints are ax = −16 m/s2 + [0.5 m sin θ ]αBC − [18 m/s2 ] cos θ ay = [0.5 m cos θ ]αBC + [18 m/s2 ] sin θ

The dynamic equations



Fx : −Bx = max

F 

y

: B y = may

MG : B x (0.5 m ) sin θ − By (0.5 m ) cos θ =

Solving we find

αBC = 17.0 rad/s2

1 12

m(1.0 m )2 αBC

CC W


1

Problem 18.65 Bars OQ and PQ each weigh 6 lb. The weigh t of the coll ar P and friction between the collar and the horizontal bar are negligible. If the system is released from rest with θ = 45◦ , what are the angular accelerations of the two bars?

Q

2 ft

2 ft

θ

O P

Solution: Let α OQ and α P Q be the clockwise angular acceleration

y

of bar OQ and the counterclockwise angular acceleration of bar PQ. The acceleration of Q is

aQ = a0 + α0Q × rQ/0 =

  i 0  2cos45 ◦

j 0

k −αOQ

2 sin 45◦

0

  

Q α α

PQ

OQ

G

O

45°

P

Qy

◦ ◦ = 2αOQ sin45 i − 2αOQ cos45 j.

Qx

x

Qy Qx

The acceleration of P is O

lb 6

lb 6

aP = aQ + αP Q × rP /Q

N

  i aP i = 2αOQ sin45 ◦ i − 2αOQ cos45 ◦ j +  0  2cos45 ◦

j 0 2 sin 45◦

Equating i and j components, ◦

aP = 2αOQ sin45 − 2αP Q sin45

◦

0 = −2αOQ cos45 ◦ + 2αP Q cos45 ◦

 k  αP Q  . 0 

From the diagrams: The equation of angular motion of bar OQ is



Qx (2sin45 ◦ ) − Qy (2cos45 ◦ ) + 6cos45

1 (6/32.2)(2)2 αOQ 3

◦

=

M0 = I0 αOQ :

(1)

The equations of motion of bar PQ are

(2).



Fx = −Qx = (6/32.2)aGx

(6)



Fy = N − Qy − 6 = (6/32.2)aGy

(7)



M = (N + Qy + Qx )(cos45 ◦ ) = 1 (6/32.2)(2)2 αP Q 12

(8).

The acceleration of the center of mass of bar

PQ is

(5).

◦

aG = aQ + αP Q × rG/Q = 2αOQ sin45 i

− 2αOQ cos45

◦

  i j +  0  cos45 ◦

j 0 ◦ − sin45

  αP Q  . 0  k

Solving Equations (1)–(8), we obtain α OQ = αP Q = 6.83 rad/s2

Hence, aGx = 2αOQ sin45 ◦ + αP Q sin45 ◦ ◦

aGy = −2αOQ cos45 + αP Q cos45

(3); ◦

(4).


1

Problem 18.66

In Problem 18.65, what are the angular accelerations of the two bars if the collar P weighs 2 lb? Solution: In the solution of Problem 18.65, the free body diagram of bar PQ has a horizontal component P to the left where P is the force exerted on the bar by the collar. Equations (6) and (8) become





◦

M = (N − P + Qy + Qx )(cos45 ) =

1 12

(6/32.2)(2)2 αP Q

and the equation of motion for the collar is P = (2/32.2)aP solving equations (1– 9), we obtain α OQ = αP Q = 4.88 rad/s2 .

Fx = −Qx − P = (6/32.2)aGx


1

Problem 18.67 The 4-kg slender bar is pinned to 2kg sliders at A and B . If friction is negligible and the system is released from rest in the position shown, what is the angular acceleration of the bar at that instant?

A

Solution: Express the acceleration of B in terms of the acceleration of A , a B = aA + αAB × rB/A : 1.2 m



 i aB cos45 ◦ i − aB sin45 ◦ j = −aA j +  0  0.5 or and



j 0 −1.2

k  αAB  , 0 

◦

(1);

◦

(2).

aB cos45 = 1.2αAB ,

B

− aB sin45 = −aA + 0.5αAB ,

We express the acceleration of G in terms of the acceleration of aG = aA + αAB × rG/A:



 i aG = aGx i + aGy j = −aA j +  0  0.25 or

45°

j 0 −0.6

A,

0.5 m



k  αAB  , 0 

A aA

aGx = 0.6αAB ,

(3);

aGy = −aA + 0.25αAB ,

(4);

G

y

and

αAB

B

The free body diagrams are as shown. The equations of motion are Slider A :

x aB

Ay

N − Ax = 0

(5),

and

(6);

(2)(9.81) + Ay = 2aA ,

Ax B By

Slider B :

P − [Bx + By + (2)(9.81)]cos45

◦

= 0,

(7);

(4)(9.81)

( ) 8 ;

Ay

Bx ◦

and [ (2)(9.81) − Bx + By ]cos45 Bar: and

= 2aB ,

(9),

Ax + Bk = 4aGx Ay + By − (4)(9.81) = 4aGy

(L/2)[(Bx − Ax ) cos β + (By − Ay ) sin β ] =

Ax

(10)

1 12

(4)L2 αAB

N

(2)(9.81) By

(11),

P

Bx

where

L=



(0.5)2 + (1.2)2 m

(2)(9.81) and

β arctan(0.5/1.2) = 22.6◦ .

Solving Equations (1)–(11), we obtain α AB = 5.18 rad/s2 .


1

Problem 18.68 The mass of the slender bar of m and the mass of the homogeneous disk is 4 m. The system is released form rest in the position shown. If the disk rolls and the friction between the bar and the horizontal surface is negligible, show that the disk’s angular acceleration is α 6g/ 95R counterclockwise.

R

=

2R

√

Solution: For the bar : The length of the bar is L 5R . Apply Newton’s second law to the free body diagram of the bar: Bx maGx , By NA mg maGy , where aGx , aGy are the accelerations of the center of mass of the bar. The moment about the bar center of mass is

=

+

−

By

=

=

Bx

4 mg

By

Bx

NA

f RBy

− RNA − R2 Bx = IB αAB .

ND

mg

For the disk:

Apply Newton’s second law and the equation of angular Bx motion to the free body diagram of the disk. f 4maDx , N D 4mg By 0, R By Rf ID αD

−

=

+

−

=

=

−

From kinematics: Since the syst em is released from rest, ωAB ωD 0. The acceleration of the center of the disk is aD RαD i. The acceleration of point B in terms of the acceleration of the center of the disk is

=

aB

=

=−

= aD +

αD



× rB/D = aD + 

i 0 R

−

j 0 0

k αD 0

 =−

RαD i

− RαD j.

The accel eration of the center of mass of the bar in terms of the acceleration of B is

aG

= aB +

αAB

2 × rG/B − ωAB rG/B = aB

  +  −

i 0

R

j 0 R

−2

k αAB

0

  

aG

= −R



2

αD

i



= aA + aAB × rG/A = aA = aA −

2

f

(5)

ND

(6)

RBy

− RNA − R2 Bx = − I2B αD ,

− Bx = −4RmαD , − 4mg − By = 0, + Rf = ID αD .

From (1), (2), and (3)

By

= mg − 2



9mR 16

+ 4IRB



αD .

From (1), (4) and (6),

By

αD

− αAB i − R(αD + αAB ) j. 2

RαAB

(4)

=

− RαAB j,

The accel eration of the center of mass of the bar in terms of the acceleration of A is

aG

RBy



ID R

+ 21Rm 4



αD .

Equate the expressions for B y and reduce to obtain

RαAB

= aB +

(3)

i

  + 

i 0

R

j 0 R

2

k αAB

0

  

+ RαAB j.

=

mg

  2

1 93Rm 16

+ IRD + 4IRB



.

For a homogenous cylinder of mass 4 m, ID bar of mass m about the center of mass,

IB

= 2R2 m. For a slender

= 121 mL2 = 125 mR2 .

Substitute and reduce:

aD

= 956gR

.

= aA i. Equateα the expressions D for aG , separate components and solve: αAB = − . Substitute to 2 5R R obtain a Gx = − αD , a Gy = − αD . Collect the results: 4 2 From the constraint on the motion, a A

(1)

5Rm

Bx

=− (2)

By

αD ,

4

+ NA − mg = − Rm αD , 2



1

Problem 18.69 Bar AB rotates in the horizontal plane with a constant angular velocity of 10 rad/s in the counterclockwise direction. The masses of the slender bars BC and CD are 3 kg and 4.5 kg, respectively. Determine the x and y components of the forces exerted on bar BC by the pins at B and C at the instant shown.

y C

B

0.2 m

10 rad/s

D x

A

0.2 m

Solution: First let’s do the kinematics

0.2 m

0.2 m

Velocity vB

= vA +

ωAB

B

× rB/A

= −(2 m/s )i = vB +

ωBC

C

0.2 m

= 0 + (10 rad/s)k × (0.2 m )j vC

G

10 rad/s

× rC/B

A

D

0.4 m

= −(2 m/s )i + ωBC k × (0.2 m )i = −(2 m/s )i + (0.2 m )ωBC j vD

= vC +

ωCD

Cx

× rD/C

= −(2 m/s )i + (0.2 m )ωBC j + ωCD k × (0.2 m )(i − j)

Bx

= (−[2 m/s] + [0.2 m] ωCD )i + (0.2 m )(ωBC + ωCD )j Since D is pinned we find ω CD = 10 rad/s, ωBC = − 10 rad/s

Cy

By

Cy

Dx

Acceleration aB

aA

αAB

rB/A

ωAB 2 rB/A

D

= + × − = 0 + 0 − (10 rad/s)2 (0.2 m )j = −(20 m/s2 )j aC

= aB +

αBC

y

The FBDs

× rC/B − ωBC 2 rC/B

The dynamics

2

2

= −(20 m/s )j + αBC k × (0.2 m )i − (−10 rad/s) (0.2 m )i = −(20 m/s2 )i + ([0.2 m]αBC − 20 m/s 2 ) j aD

= aC +

αCD

× rD/C − ωCD 2 rD/C

= −(20 m/s2 )i + ([0.2 m]αBC − 20 m/s 2 ) j + αCD k × (0.2 m )(i − j) − (10 rad/s)2 (0.2 m )(i − j) = (−40 m/s2 + [0.2 m]αCD )i + ([0.2 m][αBC + αBC ])j Since D is pinned we find α BC = −200 rad/s2 , αCD = 200 rad/s2 Now find the accelerations of the center of mass α

aG



FBC x : B x



+ Cx = (3 kg)(−10 m/s 2 )

FBCy : B y

+ Cy = (3 kg)(−40 m/s 2 )



MG1 : (C y

− By )(0.1 m ) = 121 (3 kg)(0.2 m )2 (−200 rad/s2 )



MD : C x (0.2 m )

+ Cy (0.2 m ) = 13 (4.5 kg)(

√

2[0.2 m])2 (200 rad/s2 )

Solving we find Bx Cx

= −220 N , By = −50 N = 190 N , Cy = −70 N

G.

2

= aB + BC × rG1/B − ωBC rG1/B = −(20 m/s 2 )j + (−200 rad/s2 )k × (0.1 m )i − (−10 rad/s)2 (0.1 m )i = (−10i − 40j)m/s2



1

Problem 18.70 The 2-kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar A slides on the smooth bar. At the instant shown, r = 1.2 m, ω = 0.4 rad/s, and the colla r is sliding outward at 0.5 m/s relative to the bar. If you neglect the moment of inertia of the collar (that is, treat the collar as a particle), what is the bar’s angular acceleration?

ω

A

Strategy: Draw individual free-body diagrams of the bar and collar and write Newton’s second law for the collar in terms of polar coordinates.

r

2m

Solution: Diagrams of the bar and collar show ing the force they

eθ

exert on each other in the horizontal plane are: the bar’s equation of angular motion is



M0 = I0 α:

− Nr =

1 (2)(2)2 α 3

r

(1)

In polar coordinates, Newton’s second law for the collar is



F = ma:

N eθ = m



2

d r dt 2

− rω 2

N

  er +

rα + 2

dr dt

e

r

N

O

 

ω eθ .

Equating e θ components,



N = m rα + 2

dr dt

ω



= (6)[rα + 2(0.5)(0.4)]

(2).

Solving Equations (1) and (2) with r = 1.2 m gives α = −0.255 rad/s2


1

Problem 18.71

In Problem 18.70, the moment of iner2 tia of the collar about its center of mass is 0.2 kg-m . Determine the angular acceleration of the bar, and compare your answer with the answer to Problem 18.70. Solution: Let C be the couple the collar and bar exert on each other: The bar’s equation of angular motion is



M0 = I0 α:

− Nr − C =

1 (2)(2)2 α 3

(1).

The collar’s equation of angular motion is



M = I α:

C = 0.2α

(2).

From the solution of Problem 18.70, the second law for the collar is N = (6)[rα + 2(0.5)(0.4)]

eθ

component of Newton’s

(3)

Solving Equations (1)–(3) with r = 1.2 m gives α = −0.250 rad/s2 .


1

Problem 18.72

The 3-Mg rocket is accelerating upward at 2 g  s. If you model it as a homogenous bar, what is the magnitude of the axial force P at the midpoint? 6m

Solution: At the midpoint, the mass abov e the midpoint is 3 2

m 2

=

Mg = 1500 kg. Apply Newton’s second law to the free body diagram: P−

m 2

g=

m 2

mg

a,

where a = 2g . Rearrange: P =

m 2

(g + 2g) = 1500(3)(9.81) = 44,145 N ,

P

P = 44.1 kN


1

Problem 18.73 The 20-k g slender bar is attached to a vertical shaft at A and rotates in the horizontal plane with a constant angular velocity of 10 rad/s. What is the axial force P at the bar’s midpoint?

x

y 10 rad/s 1m

A

Solution: The mass of the outer half of the bar is Apply Newton’s law to the free body diagram −P =

m

m 2

= 10 kg.

aCM , where

2 aCM is the acceleration of the center of mass of the outer half of the bar. From kinematics

aCM = ω × (ω × rCM/A ) = ω

i 0 ×  3 4

aCM = −

3ω 2 4

j 0

k ω

0

0

 i  =  0   0

j 0 3ω

4

k ω

0

  , 

i = −75i m/s 2 ,

from which P = (10)(75) = 750 N


1

Problem 18.74 The 20-kg slender bar is attached to a vertical shaft at A and rotates in the horizontal plane with a constant angular velocity of 10 rad/s. Draw a graph of the axial force in the bar as a function of x . Solution: The vector location of the center of mass of the portion of the bar outboard of x is rCM/A =



1−x 2



+ x i ( m) =



1+x 2



i ( m).

Apply Newton’s law to the outboard part of the bar: − P = m0 aCM . The mass of the outboard portion is

m0 = m



1−x 1



= m(1 − x) kg,

from which −P = m(1 − x)a CM . From kinematics, the acceleration

1000 900 800 700 N 600 , e c 500 r o F 400 300 200 100 0

Axial Force vs x

Length = 1 m ω = 10 rad/s

0 .1 .2 .3 .4 . 5 .6 .7 . 8 .9 1 x

of the center of mass of the outboard part of the bar is

aCM = ω × (ω × rCM/A ) = ω ×

from which

aCM =

    i 0

0

P =

mω

2

   

j 0 1+x

2

i 0 1+x

2



k ω



ω

0

j 0

k ω

0

0

    =−

  

, meters

,

1+x 2



ω 2 i,

2

(1 − x)( 1 + x) = 1000(1 − x 2 ) N .

The graph is shown.


1

Problem 18.75

The 100-lb slender bar AB has a builtin support at A. The y axis points upward. Determine the magnitudes of the shear force and bending moment at the bar’s midpoint if (a) the support is stationary and (b) the support is accelerating upward at 10 ft/s 2 .

y 20 lb

A

B x 4 ft

Solution:

(a)

The shea r at the midpoint is, by definition, the tot al load to the right. Thus V = (50 + 20) = 70 lb. The moment is M = −(50)(1) − 20(2) = −90 ft-lb. Check: The density per unit length is 25 lb/ft. The shear distribution is V(x) = 120 − 25x , from which V (2) = 70 lb. check. The moment is M(x) =



V(x) + C = 120x −

25 2

v

P

20 lb

x M

50 lb 1 ft

x 2 + C.

The constant of integration is found from M (4) = 0, from which C = −280, from which M (2) = −90 lb-ft. check. (b)

y

Assume that flexure of the bar as point A accelerates is negligible. 25 = 0.778 slug/ft. The shear is g

V = 20 + 50 +

1 ft

50 g

ay = 70 + 15.54 = 85.54 lb .

The moment is

The mass density is

M = −20(2) − 25(2) −

50 g

(1)ay = −90 − 15.54

V(x) = 120 + (0.778(10)(4)) − (25 + 0.778ay )x = −105.54 lb-ft.

check.

= 151.1 − 32.77x,

from which V (2) = 85.54 lb. The moment distribution is M(x) =



V(x)dx + C = 151.1x −

32.77 2

x 2 + C.

The constant of integration is determined from M (4) = 0, from which C = −342.2, and M (2) = −105.5 lb-ft. Check: The load to the right of the midpoint is


1

Problem 18.76

For the bar in Problem 18.75, draw the shear force and bending moment diagrams for the two cases. Solution: Use the solution to Problem 18.75: Shear and Moment Diagrams

Shear and Moment Diagrams

(a)

The shear and moment distributions are V (x) = 120 − 25x lb , M(x) = 120x −

(b)

25 2

x 2 − 280 lb-ft.

The shear and moment distributions are

200 150 100 50 0 –50 –100 –150 –200 –250 –300 –350

shear, lb

moment, lb -ft

01234 , ft (a)

200 150 100 50 0 –50 –100 –150 –200 –250 –300 –350

shear, lb

moment, lb - ft

01234 x

x

, ft

(b)

V (x) = 151.1 − 32.77x lb , M(x) = 151.1x −

32.77

x 2 − 342.2 lb-ft.

2 The graphs are shown.


1

Problem 18.77 The 18-kg ladder is held in equilibrium in the position shown by the force F . Neglect friction and model the ladder as a slender bar.

(a) (b)

What is the axial force, shear force, and bend ing moment at the ladder’s midpoint? If the force F is suddenly removed, what are the axial force, shear force, and bending moment at the ladder’s midpoint at that instant?

y

30° 4m

Solution: The strategy is to solve for the reactions at the surfaces,

F

and from this solution determine the axial force, shear force, and bending moment at the ladder midpoint, for the static case. The process is repeated for the dynamic case. (a) Static case reactions : Choose a coordinate system with the srcin at O and the x parallel to the floor. From geometry and the free body diagram of the ladder,

x

= iL sin θ, rA = jL cos θ, rG = L2 (i sin θ + j cos θ ),

rB

where θ 30◦ . Apply the static equilibrium conditions to the free body diagram: F A 0, B W 0. The moment about the center of mass of the ladder is

=



MG



MG

− + =

− =

= rB/G × (−F i + B j) + rA/G × Ai,



i

j

k 0 0

= L2  sin θ − cos θ −F B

   + −

i sin θ A

L

2

j cos θ 0

k 0 0

 

= L2 (B sin θ − (A + F ) cos θ) k = 0. Substitute numerical values and solve: B 176.58 N, A 50.97 N, F 50.97 N. Static case axial force, shear force, and bending moment at midpoint: Consider the lower half of the ladder, and note that from the definition of the bending moment , M bend M . Use the definitions and coordinate system for the static case reactions given above. Apply the equilibrium conditions to the free body diagram:

=

=

=

=−

(1) P cos θ

(2)

+B −

  W

2

cos θ

+ V sin θ = 0,

− P sin θ + V cos θ − F = 0,

(3) M

−

  L

4

(V

+ F cos θ ) +

from which P 101.9 N, V M 44.15 N-m.

− =

=−

from which ,

  L

4

B sin θ

= 0,

= 0M = −44.15 N-m, Mbend =

(b) Dynamic case; the reactions : The force F is zero. From the free body diagram, the application of Newton’s second law and the equation of angular motion for the dynamic case yields the three equations: A

= maGx , B − W = maGy , L2 (B sin θ − A cos θ ) = IG α,



1

where a Gx , aGy are the accelerations of the center of mass. From the constraint on the motion, the acceleration of points A and B are aA

A θ

= aA j (m/s2 ),

= aB i ( m/s2 ), ◦ where θ = 30 . From kinematics, the acceleration of G in terms

W

aB

F

O

of the acceleration at A is

aG

= aA + α × rG/A = aA + L2

 

i 0

j 0

k α 0

+ sin θ − cos θ

 

B

P

M V

L

= aA j + 2 (α cos θ i + α sin θ j) (m/s2 ), from which a Gx

= L2 α cos θ = √3α m/s2 .

The acceleration of the point G in terms of the acceleration at B is

aG



= aB + α × rG/B = aB i + L2 

−

i 0 sin θ

+

j 0 cos θ

k α 0

 

L

= aB i − 2 (α cos θ i + α sin θ j),

where a G is the acceleration of the center of mass of the lower half of the ladder, from which

 aGy

=−

aG

=

3L 4

sin θ i

+

  L

4

rG/A

=

rG/B

=

4

= −α m/s2 .



i 0 3sin θ

−

j 0 3cos θ

k α 0

 

= aA j + L4 (3α cos θ i + 3α sin θ j) √ 3 3

3α cos θ

=

=

α m/s 2 .

4

Apply Newton’s second law and the equation of angular motion to the free body diagram of the lower half of the ladder (see diagram in part (a), with F 0) and use the kinematic relations to obtain:

 = − m α, + B − W2 + V sin θ = m2 aGy 4 √ m  (2 ) V cos θ − P sin θ = a = 3 4 3 mα 2 Gx

(i sin θ

  L

4

− j cos θ ) (m),

( i sin θ

−

+ j cos θ) (m).



= aB + α × rG/B = aB + L4 

−

i 0 sin θ

L

= aB i − 4 α(i cos θ + j sin θ ) (m/s2 ).

c 2005 Pearson Educat ion, Inc., Upper 2

α sin θ

(1 ) P cos θ

The moment about the center of mass is

From kinematics: the acceleration of the midpoint of the lower half in terms of the acceleration at B is

aG

2

=

cos θ j ( m),

from which 3L

L

= aA + α × rG/B = L4 

 aGx

rG

 

aGy

Axial force, sh ear force, andladder. bendingThe moment midpoint sider the lower half of the vectoratlocat ion :ofConthe center of mass is

 

B

The acceleration of

= − L2 α sin θ = −α. Substitute into the expressions for Newton’s laws √ to obtain the three equations in three unknowns: A = m 3α , B − W = −mα , B sin θ − IG α A cos θ = , where IG = (1/12)mL2 = 24 kg-m2 . Solve: 2 B = 143.5 N, A = 57.35 N, α = 1.84 rad/s2 . Check: From 3g Example 18.4, α = sin θ = 1.84 rad/s2 , check. 2L from which

F

W 2

j 0 cos θ

k α 0

 

(3 )

+ M − (1)V + (1)B sin θ = IG α,

where

 IG

=

   1

m

12

2

(22 )

= 3 kg-m2 .

Solve: P 76.46 N, V 5.518 N, M 60.70 N-m . From the definition of the bending moment, Mbend M 60.70 N-m.

=−

=

=−

=− =

Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem 18.78 For the ladder in Problem 18.77, draw the shear force and bending moment diagrams for the two cases. Solution: Choose a coordinate system as in the solution to Problem 18.77. Use the solution to Problem 18.77 for the reactions at A and B for both cases. Cut the bar at a distance x from the pointA , and consider the lower half of the bar, (as shown in the free body diagram), noting that by the definition of the bending moment, Mbend = −M . The mass density of the ladder per unit length is m L

18

=

4

(a)

m L

rG = x sin θ i + (L − x) cos θ j (m),

rG/A =

The mass of the lower part of the ladder is mx =

Dynamic case: Use the solution in Problem 18.77 for the reactions: B = 143.5 N , α = 1.84 rad/s2 . Use the free body diagram and results of (a). The vector distance to the center of mass of the lower part from A is

from which

= 4.5 kg/m.

 

(b)

(L − x) kg.



2

and r G/B = −

Static Case: In the solution to Problem 18.77 the reactions B, F were B = 176.58 N , F = 50.97 N. Apply the static equilibrium conditions to the free body diagram to obtain:



L+x



sin θ i −

L−x

2





L+x

2

cos θ j (m)

2



j 0 − cos θ

k α 0

sin θ i +





L−x

cos θ j (m).

From kinematics: aG = aA + a × rG/A ,

(1) B −

W (L − x) L

+ V sin θ + P cos θ = 0, aG = aA +

(2) cos θ − P sin θ − F = 0.

The moment about the center of mass of the lower part of the ladder is (3) M −



L−x

2



(V + F cos θ ) + B



L−x

2



= aA j +

sin θ = 0.

P = W

 

V = W

L−x L L−x L

aGx =

WL

2



L−x L





L+x

2



x−

x2

2L

sin θ



= aB i −

L−x L

i 0 sin θ

,

(α cos θ i + α sin θ j) (m/s2 )



α cos θ.

(L − x)

2

(L − x)

2

 

i 0

j 0

k α

− sin θ

cos θ

0

 

,

(i cos θ + j sin θ) (m/s2 ),

WL

2

+

aGy = −

(L − x)

2

α sin θ.

Apply Newton’s second law and the equation of angular motion to the free body diagram, and substitute kinematic results to obtain B−

W (L − x) L

+ V sin θ + P cos θ = m



2



 

from which

− (B sin θ − F cos θ)x + C.

(B sin θ − F cos θ )L. Substitute:





2

At x = L the bending moment is zero, from which C = −

2

2

aG = aB i +

− B sin θ + F cos θ,

V dx + C = W

WL

L+x

aG = aB + a × rG/B

− B cos θ − F sin θ,

+ (B sin θ − F cos θ)(L − x)

Mbend = −



 

The graph is shown. [Check: By definition, the bending moment is the integral of the shear,



2

  

 

and M bend = −M = −

Mbend =

L+x

from which

These three equations have the solutions:

 



sin θ + (B sin θ − F cos θ)(L − x).

= −m

L−x L





aGy

(L − x) 2

2L



α sin θ.

check. At x = 0, M bend = 0, from which − W L sin θ + (B sin θ − F cos θ)L = 0 2

check. ]


1

V cos θ − P sin θ = m



L−x

=m



L−x

M+



L−x

2



L



aGx



L+x

2



α cos θ.

(−V + B sin θ ) = IG α N-m ,

   1

where I G =

L

m

12

L−x L



(L − x) 2 =

m(L − x) 3

12L

.

The graphs are shown.

P

M

V

L–x

W

F

(L – x) L B

Shear and bending moment 80 60

Moment , N-m

40 20 0 –20

Shear, N

–40 –60

0

.5

.1 1.5 2 x, m

2.5

3

3.5

4

Shear and bending moment 80 Bending Moment, N-m

60 40 20 0 –20

Shear, N –40 –60 –00

2

Dynamic Case 0

.5

1

1.5

2

2.5 x, m

3

3.5

4


Problem 18.79

Continue the calculations presented in Example 18.8, using t = 0.1 s, and determine the ladder’s angular position and angular velocity at t = 0.6 s and t = 0.7 s. Solution: The time was expressed as an array (list) such thatt [i ] = t [1] + (i − 1)t( 1 ≤ i ≤ 8), t [1] = 0, t = 0.1 s. The first values in the arrays for θ , ω and α are θ [1] = 5◦ = 0.0873 rad, ω [1] = 0, α [1] = 3g sin(θ [1]) = 0.3206 rad/s2 . The algorithm for integration is 2L

For i = 2 to i = 8,

4m

θ [i ] = θ [i − 1] + ω[i − 1] dt , α [i − 1] =

3g 2L

sin(θ [i − 1]),

5

°

ω[i ] = ω[i − 1] + α [i − 1] dt ,

i. The values are tabulated. The first values agree with Example 18.10, as a check. Next

t, s 0 .1 .2 .3 .4 .5 .6 .7

θ , rad

.0873 .0873 .0905 .0969 .1066 .1199 .1371 .1587

ω, rad/s

0 .0321 .0641 .0974 .1329 .1721 .2161 .2664

α , rad/s2

.3206 .3206 .3324 .3559 .3915 .4401 .5029 .5815


1

Problem 18.80

The moment of inertia of the helicopter’s rotor is 400 slug-ft2 . It starts from rest at t 0, the engine exerts a constant torque of 500 ft-lb, and aero2 dynamic drag exerts a torque of magnitude 20 ω ft-lb, where ω is the rotor’s angular velocity in radians per second. Using t 0.2 s, determine the rotor’s angular position and angular velocity for the first five time steps. Compare your results for the angular velocity with the closed-form solution.

=

=

Solution: The angular acceleration is obtained from the equation of angular motion for the rotor, I α T 20ω2 . For convenience, dω denote b 20. By definition, α , from which, separating varidt dω dt ables, . Make the reasonable assumption that T > bω 2 T bω2 I over the time interval of interest . Integrate:

is the closed form solution for the angular position. Substitute numerical values: ω 5 tanh(0.25t),θ 20ln (cosh(0.25t)) . Check:

√1

For the numerical integration the time is in an array t [i ] t [1] 6), where t 0.2 s, t [1] 0. The initial values (i 1)t( 1 i are θ [1] 0, ω[1] 0. The numerical integration uses the algorithm: For i 2 to i 6,

= −

=

=

=

−

tanh−1

bT

  = bω T

t I

+ C,

tanh

b

1

T

ω

√

= 5 tanh(0.25t).

=

t

I

+ C1 , α [i

from which

ω

=



T b

When t

ω

=



tanh

√  bT

I

t

tanh

b

bT

t

which is the closed form solution for the angular velocity. Although not required by the problem , the closed form for the angular position

t, s

is a straightforward integration: By definition,

0 0 0.2 0.4 0.6 0.8 1 1.2

ω

=

dθ dt

=



T b

tanh

√  bT

I

t

.

Integrate:



θ

=

θ

= Ib ln

  √   √ 

ω dt

T

+C =

cosh

b

bT

I

tanh

bT

I

t

dt

θ

= Ib ln

= √

=

check.

=

=

=

=

+

= θ [i − 1] + ω[i − 1] dt , = ω[i − 1] + α[i − 1] dt ,

ωCF

0 0.2498 0.4983 0.7444 0.9869 1.225 1.457

0

ωEuler

θCF

0 0.25 0.4994 0.7469 0.9913 1.231 1.466

0.025 0.0998 0.2242 0.3974 0.6186 0.8868

θEuler

0 0.05 0.1499 0.2993 0.4975 0.7438

+ C,

+C

t

where C is the constant of integration. When t ln(1) 0, thus θ 0 from which C 0.

=

≤ ≤

cosh(0.25t)

Next i. The results are tabulated. The first column is time, in seconds. The second column is the closed form solution for ω , rad/s. The third column is the value of ω, rad/s obtained from numerical integration. The fourth column is the closed form value of θ , rads. The fifth column is the value of θ , rads, obtained from the numerical integration. Note that the latter values show poor agreement for the first five steps. (The agreement improves as the number of steps increases.)

  I

=

20(0.25) sinh(0.25t )

− 1] = 1.25 − 0.05ω2 [i − 1], ω[i ]

+ C2 .

= 0, ω = 0, from which C 2 = 0. √

T

= dt 20ln (cosh(0.25t)) =

θ [i ]

bT

=

dt

=

d

−

where C is a constant of integration. Rearrange:

  −

=

dθ

=

= 0, cosh (0) = 1,

   cosh

bT

I

t



1

Problem 18.81

In Problem 18.80, draw a graph of the rotor’s angular velocity as a function of time from t = 0 to t = 10 s, comparing the closed-form solution, the numerical solution using t = 1.0 s and the numerical solution using t = 0.2 s. Solution: The graphs are shown. The difference between the closed form and the numerical solution is difficult to see on the scale of the graph for t = 0.2 s, but the closed form is slightly higher than the numerical solution over the entire range. The non-agreement is easy to see for t = 1 s.

r a d s / s

Rotor angular velocity vs time 5 4.5 Closed form 4 Numerical 3.5 3 ∆t = 0.2 s 2.5 2 1.5 1 .5 0 0 2 4 6 8 10 time, s

r a d s / s

5 4.5 4 3.5 3 2.5 2 1.5 1 .5 0

Rotor angular velocity vs time

Closed form

Numerical

∆t

0

2

4

6 8 time, s

=1s 10


1

Problem 18.82

The slender 10-kg bar is released from rest in the horizontal position shown. Using t = 0.1 s, determine the bar’s angular position and angular velocity for the first five time steps.

1m

Solution: From the equation of angular motion, the moment about the pinned support at A is

WG

  L

2

where

1m

cos θ = IA α,

IA =

m

3

Ay

L2 ,

Ax θ

WG = mg,

from which, by definition, α =

dω dt

=

3g 2L

cos θ .

W

Substitute numerical values: dω dt

= 14.715cos θ.

The algorithm for numerical integration is θ [1] = 0, ω [1] = 0, For i = 2 to i = 6, θ [i ] = θ [i − 1] + ω[i − 1]dt , α [i − 1] = 14.715cos (θ [i − 1]), ω[i ] = ω[i − 1] + α [i − 1]dt , Next i. The results are tabulated.

t , (s)

ω, (rad/s)

θ , (rad)

0 0.1 0.2

0 1.472 2.943

0 0 0.1472

0.3 0.4 0.5

4.399 5.729 6.665

0.4415 0.8813 1.454


1

Problem 18.83

In Problem 18.82, determine the bar’s angular position and angular velocity as functions of time from t = 0 to t = 0.8 s, using t = 0.1 s, t = 0.01 s, and t = 0.001 s. Draw the graphs of the angular velocity as a function of the angular position for these three cases, and compare them with the graph of the closed-form solution for the angular velocity as a function of the angular position. Solution: From the solution to Problem 18.82, the angular acceleration is α=

3g 2L

cos θ.

Angular Velocity vs angle

ω,


6

6

5

5

4

4

rads/s 3

Use the chain rule and the definition of the angular velocity to obtain

3 Closed form

2 1

dω

dω dθ

dω

3g

= 0.1 s

Numerical Solution

1

0

α = dt = dθ dt = ω dθ = 2L cos θ.

∆t

2

Solution

0

.2

.4

.6 .8 θ , rads

1

0

0 .2

.4

.6

.8 θ,

1 rads

Separate variables and integrate. Angular Velocity vs angle 2

ω =

3g L

sin θ + C.

For θ = 0, ω = 0, from which C = 0, and the closed form solution is

ω=



ω,


6

6

5

5

4

4

rads/s 3

∆t

3

= 0.01 s

2

3g L

sin θ ,

where the positive sign has been chosen from physical reasoning (the bar is swinging counterclockwise). The algorithm for numerical integration is that given in the solution to Problem 18.82.

1 0

∆t

= 0.001 s

2 1

Numerical Solution 0

.2

.4

.6 θ , rads

.8

1

0

0 .2

Numerical Solution .4

.6

.8 θ,

1 rads


1

Problem 18.84

In Problem 18.82, suppose that the bar’s pin support contains a damping device that exerts a resisting couple on the bar of magnitude cω (N-m), where ω is the angular velocity in radians per second. Using t = 0.001 s, draw graphs of the bar’s angular velocity as a function of time from t = 0 to t = 0.8 s for the cases c = 0, c = 2, c = 4, and c = 8. Solution: From the application of the equation of angular motion to the bar (see the solution to Problem 18.82)

WG

  L

2

where

o 5 m e g 4 a , 3 r

cos θ − cω = IA α,

WG = mg, IA =

from which

m

3

dω dt

L2 kg-m2 ,

=

3g 2L

cos θ −

Response of damped bar 6

3c mL2

c

=0

c

=2

c

a 2 d / s 1

ω.

=4

c

=8

Substitute numerical values: 0 dω dt

= 14.715cos θ − 0.3cω.

The algorithm for numerical integration is: θ [1] = 0, ω[1] = 0, i = 2 to i = 6,

0

.1

.2

.3

.4

.5

.6

.7

.8

.9

time, s For

θ [i ] = θ [i − 1] + ω[i − 1] dt , α [i − 1] = 14.715cos (θ [i − 1]) − 0.3cω [i − 1], ω[i ] = ω[i − 1] + α [i − 1] dt , Next

i. The graphs are shown.


1

Problem 18.85 The 18-kg ladder is released from rest in the position shown. The floor and wall are smooth. The ladder will lose contact with the wall before it hits the floor. Using t = 0.001 s, estimate the time and the value of the angle between the wall and the ladder when this occurs. Strategy: The formulation of the problem assumes that the ladder remains in contact with the wall. For times greater than the time at which it would lose contact, the solution for the normal force exerted on the ladder by the wall will become negative. So you can determine the time at which contact is lost by determining the time at which the normal force decreases to zero.

4m

5

Solution: Choose a coordinate system with the srcin at the corner of the wall and floor, with the x axis parallel to the floor. Denote the contact point with the wall by P , the contact with the floor by N , and the center of mass by G . From Newton’s second law and the equation of angular motion, P = maGx , N − mg = may , and

N



 

L sin θ

L cos θ

−P

2

where I G =

  1

12

2



= IG α,



The acceleration of the center of mass in terms of the acceleration of point N is aG = aN +

aG = aN +

mL2 . From kinematics: The vector distance

rG/P = rG − rP =

− ω2 L

2

(i sin θ − j cos θ).

α

   

× rG/N − ω2 rG/N ,

−

−

aG = aP +

− ω2

   

i 0 L sin θ

−

2

L sin θ

2

i−

j 0 L cos θ

2

L cos θ

2

k α

0



j .

L sin θ

2

i+

2

L cos θ

2

k α

0

  



j .

aN = aN i,

from which

  

+

2

j 0 L cos θ

The constraint on the motion at N is such that

The acceleration of the center of mass in terms of the acceleration at point P is aG = aP + α × rG/P − ω2 rG/A ,

i 0 L sin θ

aGy =

L

2

(α sin θ + ω2 cos θ)

Collect results and substitute the kinematic relations to obtain: mL

(α cos θ − ω2 sin θ) ,

(1)

P =

(2)

N = mg −



2

2

(α sin θ + ω2 cos θ) ,



L cos θ −P = IG α . 2 2 Substitute (1) and (2) into (3) and reduce to obtain

(3)

N

L sin θ

mL

 

The constraint on the motion at P is such that aP = aP j,

from which

aGx =

L

2

(α cos θ − ω2 sin θ).

α=

2 mg L sin θ 4IG + mL2

=

3g 2L

sin θ.

The ladder leaves the wall when P = 0, when α cos θ = ω2 sin θ , so that (2) and (3) are not required in the numerical solution. Substitute numerical values: m = 18 kg, L = 4 m, to obtain P = 36(α cos θ − ω2 sin θ) N, α = 3.68 sin θ rad/s2 . A copy of the algorithm used in TK Solver Plusis shown. The algorithm is a called procedure, returning the time and angle at which the force exerted by the wall vanishes. These values are: t = 1.554 s and θ = 0.8455 rads = 48.44◦ . Although not required by the problem, the graph of the force P against the time is shown.


1

[Check: An analytic solution: From P = 0, α cos θ − ω2 sin θ = 0. 3g dω Substitute α = sin θ . Noting α = , use the chain rule: 2L dt dω dt

=

dω dθ

dω

=ω

dθ dt

=

dθ

3g 2L

F o r c e , N

sin θ.

Separate variables and integrate: ω2 2 = −

3g 2L

cos θ + C.

Assume that ω = 0 at θ → 0 (the ladder won’t start to fall if exactly, but we suppose that θ is very small),

θ= 0

Force exerted by wall 50 Initial values: 45 2 40 α = 0.3206 rad/s θ =5 35 ω =0 30 25 P 20 15 10 5 0 –5 0 .5 1 1.5 2 °

time, s from which

C=

from which

ω=

3g 2L

,

3g



L (1 − cos θ),

where the positive sign is taken because the ladder rotates counterclockwise. Substitute: 3g 2L

sin θ cos θ −

3g L

(1 − cos θ) sin θ = 0.

Reduce algebraically to obtain cos θ =

2 3

from which θ = 48.2◦

P

ω

θ

N W

2


Problem 18.86

A torsional spri ng exerts a counterclockwise couple k θ on the bar, where k = 20 N-m and θ is in radi ans. The 2-kg bar is 1 m long. At t = 0, the bar is released from rest in the horizontal position (θ = 0). Using t = 0.01 s, determine the bar’s angular position and angular velocity for the first five time steps.

k

θ

Solution: The bar’s equation of angular motion is

 or

M0 = I0 α :

mg(l/ 2) cos θ − kθ =

(2)(9.81)(1/2) cos θ − 20θ =

1 3

1 3

k

ml 2 α θ

(2)(1)2 α.

dw We see that α = = 14.715cos θ − 30θ . dt

mg

Initial Conditions: At t 0 = 0, θ (t0 ) = 0 and ω (t0 ) = 0. First Time Step: At t = t0 + t = 0.01 s, the angle is

θ (t0 + t) = θ( t0 ) + ω(t 0 )t = 0, and the angular velocity is

ω(t 0 + t) = ω(t 0 ) + α(t0 )t = [14.715cos (0) − 30(0)](0.01) = 0.1472 rad/s.

Second Time Step: At t = t0 + 2t = 0.02 s the angle is

θ (t0 + 2t) = θ (t0 + t) + ω(t 0 + t)t = (0.1472)(0.01) = 0.0015 rad,

and the angular velocity is

ω(t 0 + 2t) = ω(t 0 + t) + α(t0 + t)t = 0.1472 + [14.715cos (0) − 30(0)](0.01) = 0.2943 rad/s.

Continuing, we obtain

t, s

θ , rad

0.00 0.01 0.02 0.03 0.04 0.05

0.0000 0.0000 0.0015 0.0044 0.0088 0.0147

ω, rad/s 0.0000 0.1472 0.2943 0.4410 0.5868 0.7313


1

Problem 18.87

Using a numerical solution with t = 0.001 s, estimate the maximum angle θ reached by the bar in Problem 18.86 when it is released from rest in the horizontal position. At what time after release does the maximum angle occur? Solution: Carrying out the numerical solution in the manner described in the solution of Problem 18.86, the resulting graph of θ as a function of time is shown below: By examining the computer results near the maximum, we estimate that the maximum angle θ = 0.867 rad ◦ or ( 49.7 ) occurs at t = 0.524 s.

0.9 0.8 0.7 0.6 d a r , θ

0.5 0.4 0.3 0.2 0.1 0

0

0.2

0.4

0.6 t, s

0.8

1


1

Problem 18.88

The axis L0 is perpendicular to both segments of the L-shaped slender bar. The mass of the bar is 6 kg and the material is homogeneous. Use integration to determine the moment of inertia of the bar about L0 .

1m

LO

2m

Solution: Let A be the bar’s cross-sectional area. The bar’s mass is m = 6 kg = ρA( 3 m ), so ρ A = 2 kg/m.

dm dx

For the horizontal part (Fig. a),

Ih =



x 2 dm = m



=



r 2 dm = m

13 3

ρA =



26 3

dy y

x 2

x 2 ρAdx = 0

8 3

ρA =

For the vertical part (Fig. b),

Iv =

r

16 3

kg-m2 .

LO

dm

(a)

LO

(b)

1

(22 + y 2 )ρAdy 0

kg-m2 .

Therefore I0 = Ih + Iv = 14 kg-m2 .


1

Problem 18.89

Two homogenous slender bars, each of mass m and length l , are welded together to form the Tshaped object. Use integration to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars.

l

O l

Solution: Divide the object into two pieces, each corresponding to a slender bar of mass m ; the first parallel to the y -axis, the second to the x -axis. By definition

I

=



l

r 2 dm 0

+



r 2 dm. m

For the first bar, the differential mass is dm ρAdr . Assume that the second bar is very slender, so that the mass is concentrated at a distance l from O . Thus d m ρAdx , where x lies between the limits l l x . The distance to a differential dx is r l 2 x 2 . Thus 2 2 the definition becomes

=

=

− ≤ ≤ I

I

l

= ρA



= ρA

 

= ml2



r 2 dr 0

r3

3

1 3

+ ρA

l 0

+ ρA





1 2

− 2l

(l 2

=

+

+ x 2 ) dx 3

l2x

√

+ x3



1 2

− 12



+ 1 + 121 = 17 ml 2 12



1

Problem 18.90

The slender bar lies in the x −y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the z axis.

y

2m



50

x

1m

Solution: The density is ρ =

Iz =



1 m 0

+



6 kg 3m

= 2 kg/m y

ρx 2 d x

2 m

◦

◦

ρ [(1 m + s cos50 )2 + (s sin50 )2 ] d s

2m

0

Iz = 15.1 kg-m2

50°

1m

x


1

Problem 18.91

The slender bar lies in the x −y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the y axis. Solution: The density is ρ =

Iy =



1 m

6 kg 3m

= 2 kg/m y

2

ρx d x 0

+



2 m

2m

◦

ρ [(1 m + s cos50 )2 ] d s

0

50°

Iy = 12.0 kg-m2 1m

x


1

Problem 18.92 The homogeneous thin plate has mass m = 12 kg and dimensions b = 1 m and h = 2 m. Determine the mass moments of inertia of the plate about the x , y , and z axes.

y

h

Strategy: The mass moments of inertia of a thin plate of arbitrary shape are given by Eqs. (18.37)–(18.39) in terms of the moments of inertia of the cross-sectional area of the plate. You can obtain the moments of inertia of the triangular area from Appendix B.

x

b

Solution: y m = 12 kg 1 bh 2

Area =

h x

ρ = mass/Area dm = ρd A b

From Appendix B,

IxA =

1 36 1

Area =

2

bh3

IyA =

1 36

hb3

Ix =

12 36

h

b3 =

1 3

(2)(1)3

Iy = 0.667 kg-m2 (1)(2) = 1 m2 Iz = Ix + Iy

ρ = 12 kg/m2



Iy = ρI yA =

2

Iz = 2.667 + 0.667 kg-m2

ρy dA = ρ



2

y dA

Iz = 3.333 kg-m2

Ix = ρI xA , Iy = ρI yA Ix = 12

  1

36

(1)(2)3 = 2.667 kg-m2


1

Problem 18.93

The brass washer is of unifor m thickness and mass m. (a) (b)

Determine its moments of inertia about the x and z axes. Let R i = 0, and compare your results with the values given in Appendix C for a thin circular plate.

y Ro Ri x

Solution:

(a)

The area moments of inertia for a circular area are

π R4

Ix = Iy =

4

Ri

Ro

.

For the plate with a circular cutout,

Ix =

π 4

(Ro4 − Ri4)

The area mass density is cut,

m A

=

m π(R o2 − Ri2)

m A

, thus for the plate with a circular

,

from which the moments of inertia

I(x -axis) =

m(Ro4 − Ri4) 4(Ro2 − Ri2 )

I(z-axis) = 2I(x -axis) = (b)

m 2

=

m 4

(Ro2 + Ri2)

(Ro2 + Ri2).

Let R i = 0, to obtain

Ix -axis =

I(z-axis) =

m 4

Ro2 ,

m 2

Ro2 ,

which agrees with table entries.


1

Problem 18.94

The homogenous thin plate is of uniform thickness and weighs 20 lb. Determine its moment of inertia about the y axis.

y

1

2 y = 4 – – x ft

4

x

Solution: The definition of the moment of inertia is I =



r 2 dm. m

The distance from the y -axis is x , where x varies over the range m W be the area mass density. The mass −4 ≤ x ≤ 4. Let τ = = A gA W of an element y dx is d m = y d x . Substitute into the definition: gA Iy -axis =

=

W gA W gA

4

  x2

4−

−4



4x 3 3

−

x5

20



x2

4



dx

+4

= −4

W gA

[68.2667].

The area is 4

A=

 

4−

−4

x2

4





d x = 4x −

x3

12



4

= 21.333 ft2 −4

The moment of inertia about the y -axis is

I(y -axis) =

W g

(3.2) =

20 32.17

(3.2) = 1.99 slug-ft2 .


1

Problem 18.95

Determine the moment of inertia of the plate in Problem 18.94 about the x axis. Solution: The differential mass is d m =

W gA

dy dx . The distance

of a mass element from the x -axis is y , thus

I =

=

=

W gA



W

3gA W

3gA

+4

dx −4



4−

4−

x2

−4



y 2 dy

0

+4

 

x2 4

4

64x − 4x 3 +

3



dx

3 20

x5 −

x7

448



4 −4

W = 3gA [234.057].

From the solution to Problem 18.94, A = 21.333 ft 2 . Thus the moment of inertia about the x -axis is Ix -axis =

W (234.057)

3g (21.333)

=

W g

(3.657) = 2.27 slug-ft2 .


1

Problem 18.96

The mass of the object is 10 kg. Its moment of inertia about L1 is 10 kg-m 2 . What is its moment of inertia about L2 ? (The three axes are in the same plane.)

0.6 m L

0.6 m L

1

L

2

Solution: The strategy is to use the data to find the moment of inertia about L, from which the moment of inertia about L2 can be determined.

IL = −(0.6)2 (10) + 10 = 6.4 kg-m 2 ,

from which IL2 = (1.2)2 (10) + 6.4 = 20.8 kg-m 2


1

Problem 18.97

An engineer gathering data for the design of a maneuvering unit determines that the astronaut’s center of mass is at x = 1.01 m, y = 0.16 m and that her moment of inertia about the z axis is 105.6 kg-m 2 . The astronaut’s mass is 81.6 kg. What is her moment of inertia about the z  axis through her center of mass?

y

y

x

x

The distance from the z axis to the z axis is d = Solution:  x 2 + y 2 = 1.02257 m. The moment of inertia about the z  axis is Iz -axis = −d 2 m + Iz-axis = −(1.0457)(81.6) + 105.6 = 20.27 kg-m2


1

Problem 18.98

Two homogenous slender bars, each of mass m and length l , are welded together to form the T-shaped object. Use the parallel-axis theorem to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars.

l

O l

Solution: Divide the object into two pieces, each corresponding l to a bar of mass m . By definition I = 0 r 2 dm . For the first bar, the differential mass is dm = ρAdr , from which the moment of inertia about one end is



is the moment of inertia about the center of the bar. From the parallel axis theorem, the moment of inertia about O is

I0 = I1 = ρA



l

r 2 dr = ρA 0

r3

  3

l

= 0

ml 2

3

ml 2

3

+ l2m +

ml 2

12

=

17 12

ml 2

.

For the second bar

I2 = ρA



l 2 l −2

r 2 dr = ρA

r3

l 2

3

−

 

l 2

=

ml 2

12


1

Problem 18.99

Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object in Problem 18.98 about the axis through the center of mass of the object that is perpendicular to the two bars. Solution: The location of the center of mass of the obj ect is

m x

=

 l

+ lm 3 2 = l. 2m 4

Use the results of Problem 18.98 for the moment of inertia of a bar about its center. For the first bar,

I1 =

 l

2

m+

4

ml 2

12

=

7 48

ml 2 .

For the second bar,

I2 =

 l

2

m+

4

ml 2

12

=

7 48

ml 2 .

The composite:

Ic = I1 + I2 =

7 24

ml 2

Check:

Use the results of Problem 18.98:

Ic = −

 

=



3l 4

−9

8

+

2

(2m) +

17 12



17 12

ml 2 =

ml 2

7 24

ml 2 .

check .


1

Problem 18.100

The mass of the hom ogeneous slender bar is 30 kg. Determine its moment of inertia about the z axis.

y

y

x

0.8 m

x

0.6 m

Solution: The density is ρ = Iz =

1 3

(10 kg )(1.0 m )2 +

1 12

30 kg 3m

2m

= 10 kg/m

(20 kg )(2 m)2

+ (20 kg )[(1.6 m )2 + (0.8 m )2 ] Iz = 74 kg-m2


1

Problem 18.101

The mass of the hom ogeneous slender bar is 30 kg. Determine the moment of inertia of the  bar about the z axis through its center of mass. Solution: First locate the center of mass x=

y=

(10 kg )(0.3 m ) + (20 kg )(1.6 m )

30 kg (10 kg )(0.4 m ) + (20 kg )(0.8 m )

30 kg

= 1.167 m

= 0.667 m

Using the answer to 18.100 Iz = (74 kg-m2 ) − (30 kg )(1.1672 + 0.6672 )m2 Iz = 19.8 kg-m2


1

Problem 18.102

The homogeneous slender bar weighs 5 lb. Determine its moment of inertia about the z axis.

y

y

4 in

x

x

Solution: The Bar’s mass is m 5/32.2 slugs. Its length is L 82 82 π( 4) 31.9 in. The masses of the L1 L2 L3 8 parts are therefore,

√

+ + = +

+ +

M1

= LL1 m =

  

M2

= LL2 m =

√

M3

= LL3 m =

  

8

5

31.9

32.2

2(64)

31.9

=

5

32.2

4π

5

32.2

=

8 in

= 0.0390 slugs,

 

31.9

=

= 0.0551 slugs,

= 0.0612 slugs.

The center of mass of part 3 is located to the right of its center C a distance 2R/π 2(4)/π 2.55 in. The moment of inertia of part 3 about C is

=



r 2 dm m3

=

= m3 r 2 = (0.0612)(4)2 = 0.979 slug-in2 .

The moment of inertia of part 3 about the center of mass of part 3 is therefore I3 0.979 m3 (2.55)2 0.582 slug-in2 . The mome nt of inertia of the bar about the z axis is

=

I(z

1 axis)

m1 L21

− 1

m2 L22

=

I3

m3 [(8

2.55)2

=3 +3 + + + = 11.6 slug-in2 = 0.0803 slug-ft2 .

(4)2 ]

+



1

Problem 18.103 Determine the moment of inerti a of the bar in Problem 18.102 about the z axis through its center of mass. Solution: In the solution of Problem 18 .102, it is shown that the moment of inertia of the bar about the z axis is I (z axis) 11.6 slug-in2 . The x and y coordinates of the center of mass coincide with the centroid of the bar:

=

x

= x1 LL1 ++xL2 L2++Lx3 L3 1

(4)(8)

=

2

y

= y1 LL1 ++yL2 L2++Ly3 L3 1

2

3

√2 2 = 0 + (4)√82 + 82 + π( 4)(4) = 3.00 in . 8 + 8 + 8 + π( 4) The moment of inertia about the z  axis is

3

√ + (4) 82 + 82 + 8 + 2(π4) √ 8 + 82 + 82 + π( 4)





I(z axis)

π( 4)

= 6.58 in ,

= I(z axis) − (x2 + y2 )

  5

32.2

= 3.44 slug-in2 .

c 2005 Pearson Educat ion, Inc., Upper Saddle


1

Problem 18.104

The rock et is used for atmo spheric research. Its weight and its moment of inertia about the z axis through its center of mass (including its fuel) are 10,1000 lb and 10,200 slug-ft 2 , respectively. The rocket’s fuel weighs 6000 lb, its center of mass is located at x = −3 ft, y = 0, and z = 0, and the moment of inertia of the fuel about the axis through the fuel’s center of mass parallel to z axis is 2200 slug-ft 2 . When the fuel is exhausted, what is the rocke t’s momen t of inertia about the axis through its new center of mass parallel to z axis? Solution: Denote the moment of inertia of the empty rocket as I E about a center of mass x E , and the moment of inertia of the fuel as I F about a mass center x F . Using the parallel axis theorem, the moment of inertia of the filled rocket is

y

x

From which

xE = −



186.335 124.224



(−3) = 4.5 ft

is the new location of the center of mass. Substitute:

2 m + I + x2 m , IR = IE + xE E F F F

about a mass center at the srcin

(x R = 0).

IE = IR − xE 2 mE − IF − xF 2 mF .

Solve:

2 IE = IR − xE mE − IF − xF2 mF = 10200 − 2515.5 − 2200 − 1677.01 = 3810 slug-ft2

The objective is to determine values for the terms on the right from the data given. Since the filled rocket has a mass center at the srcin, the mass center of the empty rocket is found from 0 = mE xE + mF xF , from which

xE = −

  mF

mE

xF .

Using a value of g = 32 ft/s2 ,

mF =

mE =

WF g

=

6000 32.2

(WR − WF ) g

= 186.34 slug,

=

10000 − 6000 32.2

= 124.23 slug.


1

Problem 18.105

The mass of the homogeneous thin plate is 36 kg. Determine the moment of inertia of the plate about the x axis.

y

0.4m

0.4m

0.3 m

0.3 m x

Solution: Divide the plate into two areas: the rect angle 0.4 m by 0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The m mass density is ρ = . A The area is A = (0.4)(0.6) + (0.4)(0.3) = 0.36 m2 ,

from which ρ=

36 0.36

= 100 kg/m2 .

The moment of inertia about the x -axis is

Ix -axis = ρ

 1 3

(0.4)(0.63 ) + ρ

 1 3

(0.4)(0.3)3 = 3.24 kg-m2


1

Problem 18.106

Determine the moment of inerti a of the plate in Problem 18.105 about the z axis. Solution: The basic relation to use is I z-axis = Ix -axis + Iy -axis . The value of I x -axis is given in the solution of Problem 18.105. The moment of inertia about the y -axis using the same divisions as in Problem 8.105 and the parallel axis theorem is

Iy -axis = ρ

 1 3

(0.6)(0.4)3 + ρ

  1

12

(0.3)(0.4)3

+ (0.6)2 ρ( 0.3)(0.4) = 5.76 kg-m2 ,

from which Iz-axis = Ix -axis + Iy -axis = 3.24 + 5.76 = 9 kg-m 2


1

Problem 18.107

The mass of the homogeneous thin plate is 20 kg. Determine its moment of inertia about the x axis.

y

1000 mm

400 mm

400 mm

200 mm

Solution: Break the plate into the three regions shown.

y

1000

A = (0.2 m )(0.8 m ) + (0.2 m )(0.4 m )

+

ρ=

1 2

(0.4 m )(0.6 m ) = 0.36 m 2

20 kg 0.36 m2

= 55.6 kg/m2

800 200

Using the integral tables we have Ix =

1 3 +

x

200 mm

(0.2 m )(0.8 m )3 +

1 36

1 12

(0.6 m )(0.4 m )3 +

400

(0.2 m )(0.4 m )3 + (0.2 m )(0.4 m )(0.6 m )2

1 2

(0.6 m )(0.4 m )(0.667 m)2

200

x

= 0.1184 m4 Ix −axis = (55.6 kg/m2 )(0.1184 m4 ) = 6.58 kg-m2


1

Problem 18.108

The mass of the homogeneous thin plate is 20 kg. Determine its moment of inertia about the y axis. Solution: See the solution to 18.107 Iy =

1 3 +

(0.8 m )(0.2 m )3 +

1 36

1 12

(0.4 m )(0.6 m )3 +

(0.4 m )(0.2 m )3 + (0.2 m )(0.4 m )(0.3 m )2

1 2

(0.6 m )(0.4 m )(0.6 m )2

= 0.0552 m4 Iy −axis = (55.6 kg/m2 )(0.0552 m4 ) = 3.07 kg-m2


1

Problem 18.109

The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a homogeneous thin rectangular plate. The mass of the radiator is 5 slugs. Determine its moments of inertia about the x ,y , and z axes.

y

3 ft

6 ft

3 ft

2 ft x

Solution: The area is A = 9(3) = 27 ft 2 .

The mass density is

ρ=

m A

=

5 27

= 0.1852 slugs/ft2 .

The moment of inertia about the centroid of the rectangle is

Ixc = ρ

 

9(33 ) = 3.75 slug-ft2 ,

Iyc = ρ

 

3(93 ) = 33.75 slug-ft2 .

1

12 1

12

Use the parallel axis theorem: Ix -axis = ρA( 2 + 1.5)2 + Ixc = 65 slug-ft 2 , Iy -axis = ρA( 4.5 − 3)2 + Iyc = 45 slug-ft 2 . Iz-axis = Ix -axis + Iy -axis = 110 slug-ft 2


1

Problem 18.110

The mass of the homo geneous thin plate is 2 kg. Determine the moment of inertia of the plate about the axis through point O that is perpendicular to the plate.

80 mm

10 mm 30 mm O 30 mm 130 mm

Solution: By determining the moments of inertia of the area about the x and y axes, we will determine the moments of inertia of the plate about the x and y axes, then sum them to obtain the moment of inertia about the z axis, which is I 0 .

y

The areas are A1 =

2

1 (130)(80) 2

mm2 ,

1

A2 = π( 10)2 mm2 .

1 12

x

O

100 mm

Using Appendix B,

Ix =

30 mm

(130)(80)3 −



π( 10)4 + (30)2 A2



π( 10)4 + (100)2 A2



1 4

= 5.26 × 106 mm4 ,

Iy =

1 4

(80)(130)3 −



1 4

= 40.79 × 106 mm4 .

Therefore I(x

axis)

m

=

Ix = 2150 kg-mm2 ,

A1 − A2 I(y

axis)

=

m A1 − A2

Iy = 16700 kg-mm2 .

Then I(z

axis)

= I(x

I(z

axis)

= 0.0188 kg-m2 .

axis)

+ I(y

axis)

= 18850 kg-mm2 .


1

Problem 18.111 The homogeneous cone is of mass m . Determine its moment of inertia about the z axis, and compare your result with the value given in Appendix C.

y

x

Strategy: Use the same approach we used in Example 18.13 to obtain the moments of inertia of a homogeneous cylinder.

R

h

z

Solution: The differential mass m dm =

V

3m

2

 

2

π r dz = R 2 h r dz.

The moment of inertia of this disk about the z-axis is radius varies with z , r =

Iz-axis =

3mR 2 2h5



  R h

h

4

z dz = 0

1 2

mr 2 . The

z, from which

3mR 2 2h5

z5

  5

h

= 0

3mR 2 10


1

Problem 18.112

Determine the moments of inertia of the homogeneous cone in Problem 18.111 about the x and y axes, and compare your results with the values given in Appendix C. 3m m . The differential = V π R2 h element of mass is dm = ρπ r dz .. The mome nt of inertia of this elemental disk about an axis through its center of mass, parallel to the 1 x - and y -axes, is d Ix = r 2 dm . Use the parallel axis theorem, 4 Solution: The mass density is ρ = 2



Ix =

 1 4

m

r 2 dm +

Noting that r =

r 2 dm = ρ

Ix = ρ

πR 4 h4

Ix =



πR 2 h2

  3mR2 4h5

z4 dz,

 

πR 4 4h4

+

z2 dm. m

R z, then h

 

and z 2 dm = ρ



h

z4 dz . Substitute:

z4 dz + ρ

0

3m

h3

z5 5

 

πR 2 h2

 

h

=m 0



3 20

h

z4 dz,

0

R2 +

3 5

h2



= Iy .


1

Problem 18.113

The homogeneous object has the shape of a truncated cone and consists of bronze with mass density ρ = 8200 kg/m3 . Determine the moment of inertia of the object about the z axis.

y

x 60 mm

180 mm

z 180 mm

Solution: Consider an elemen t of the cone consi sting of a disk of thickness dz: We can expr ess the radius as a linear function of zr = az + b. Using the conditions that r = 0 at z = 0 and r = 0.06 m

y x

at z = 0.36 m to evaluate a and b we find that r = 0.167 z . From Appendix C, the moment of inertia of the element about the z axis is

(Iz )element =

1 2

mr 2 =

1 2

[ρ(πr 2 )dz ]r 2 =

1 2

ρπ ( 0.167z)4 dz.

r z

z dz

We integrate this result to obtain the mass moment of inertia about the z axis for the cone:

I(z axis) =

=



0.36 0.18

1 2

1 2

ρπ (0.167)4

(8200)π( 0.167)4

z5

0.36

  5

z5

  5

0.18

0.36 0.18

= 0.0116 kg-m2 .


1

Problem 18.114

Determine the moment of inerti a of the object in Problem 18.113 about the x axis. Solution: Consider the disk elemen t described in the solution to Problem 18.113. The radius of the laminate is r = 0.167z. Using Appendix C and the parallel axis theorem, the moment of inertia of the element about the x axis is

1

(Ix )element =

4

=

1 4

mr 2 + mz2 =

axis)

=

1 4

4

[ρ(πr 2 )dz ]r 2 + [ρ(πr 2 )dz ]z2

ρπ (0.167z)4 dz + ρπ (0.167z)2 z2 dz.

Integrating the result,

I(x

1

ρπ (0.167)4



0.36 0.18

z4 dz + ρπ (0.167)2



0.36

z4 dz

0.18

= 0.844 kg-m2 .


1

Problem 18.115

The homogeneous rectangular parallelepiped is of mass m . Determine its moments of inertia about the x , y , and z axes and compare your results with the values given in Appendix C.

y

a z Solution: Consider a rectangular slice normal to the x -axis of dimensions b by c and mass dm . The area dens ity of this slice is dm ρ= . The moment of inertia about the y axis of the centroid of a bc thin plate is the product of the area density and the area moment of iner1 1 tia of the plate: d Iy = ρ bc 3 , from which dI y = c2 dm . 12 12 By symmetry, the moment of inertia about the z axis is

 

dI z =

  1

12

 

b

c

By symmetry, the argument can be repeated for each coordinate, to obtain

Iy =

b2 dm.

x

m

12

(a 2 + c2 )

Iz =

m

12

(b2 + a 2 )

Since the labeling of the x - y - and z -axes is arbitrary, dI x = dI z + dI y ,

where the x -axis is normal to the area of the plate. Thus

dI x =

  1

12

b2 + c2 dm,



from which

Ix =

  1

12

(b2 + c2 )



m

dm =

m

12

(b2 + c2 ) .


1

Problem 18.116

The sphere-capped cone consists of material with density 7800 kg/m 3 . The radius R = 80 mm. Determine its moment of inertia about the x axis.

y

z

R

x 4R

Solution: Given ρ = 7800 kg/m3 , R = 0.08 m

Using the tables we have Ix =

3 10





1 2 ρ πR 2 [4R ] R 2 + 3 5



2 ρ πR 3 3



R2

Ix = 0.0535 kg-m2


1

Problem 18.117

Determine the moment of inerti a of the sphere-capped cone in Problem 18.116 about the y axis. Solution: The center of mass of a half-sphere is located a distance 3R from the geometric center of the circle. 8

Iy =





1 ρ π R 2 [4R ] 3



2 − ρ πR 3 3

3 5

[4R ]2 +

2

   3R 8

3 20

R2

2 + ρ πR 3 3

  +



2 5



2 ρ πR 3 R 2 3

4R +

3R 8

2



Iy = 2.08 kg-m2


1

Problem 18.118

The circular cylinder is made of aluminum (Al) with density 2700 kg/m 3 and iron (Fe) with density 7860 kg/m 3 . Determine its moment of inertia about the x  axis.

y

y

Al

z

Fe 600 mm 200 mm

Solution: Ix =

1 2 +

z

[(2700 kg/m2 )π( 0.1 m )2 (0.6 m )](0.1 m )2

600 mm x, x

1 2

2

2

2

[(7860 kg/m )π( 0.1 m ) (0.6 m )](0.1 m )

Ix = 0.995 kg-m2


1

Problem 18.119

Determine the moment of inerti a of the composite cylinder in Problem 18.118 about the y  axis. Solution: First locate the center of mass

x=

[(2700 kg/m3 )π( 0.1 m )2 (0.6 m )](0.3 m ) + [(7860 kg/m3 )π( 0.1 m )2 (0.6 m )](0.9 m ) (2700 kg/m3 )π( 0.1 m )2 (0.6 m ) + (7860 kg/m3 )π( 0.1 m )2 (0.6 m )

x = 0.747 m Iy = [(2700 kg/m3 )π( 0.1 m )2 (0.6 m )]



1 12

(0.6 m )2 +

1 4

(0.1 m )2



+ [(2700 kg/m3 )π( 0.1 m )2 (0.6 m )](x − 0.3 m )2

1

+ [(7680 kg/m3 )π( 0.1 m )2 (0.6 m )]

3

2



(0.6 m )2 +

12

+ [(7680 kg/m )π( 0.1 m ) (0.6 m )](0.9 m − x)

1 4

2

(0.1 m )2



Iy = 20.1 kg-m2


1

Problem 18.120

The homogeneous machine part is made of aluminum alloy with mass density ρ = 3 2800 kg/m . Determine the moment of inertia of the part about the z axis.

y

y

20 mm x

z

40 mm 120 mm 40 mm

Solution: We divide the machine part into the 3 parts shown: (The dimension into the page is 0.04 m) The masses of the parts are

y

0.12 m 1

0.08 m

m1 = (2800)(0.12)(0.08)(0.04) = 1.075 kg,

x

y

m2 = (2800) 12 π( 0.04)2 (0.04) = 0.281 kg,

C 2

m3 = (2800)π( 0.02)2 (0.04) = 0.141 kg.

y

+

0.12

0.12 m x –

0.04 m

3

x 0.02 m

m

Using Appendix C and the parallel axis theorem the moment of inertia of part 1 about the z axis is

I(z

axis)1

=

1 12

m1 [(0.08)2 + (0.12)2 ] + m1 (0.06)2

= 0.00573 kg-m2 .

The moment of inertia of part 2 about the axis through the center C that is parallel to the z axis is 1 m R2 2 2

=

1 m (0.04)2 2 2

The distance along the x axis from C to the center of mass of part 2 is 4(0.04)/(3π ) = 0.0170 m. Therefore, the moment of inertia of part 2 about the z axis through its center of mass that is parallel to the axis is 1 2 2 m2 (0.04)

2

2

− m2 (0.0170) = 0.000144 kg-m .

Using this result, the moment of inertia of part 2 about the z axis is

I(z

axis)2

= 0.000144 + m2 (0.12 + 0.017)2 = 0.00544 kg-m2 .

The moment of inertia of the material that would occupy the hole 3 about the z axis is

I(z

axis)3

=

1 m (0.02)2 2 3

+ m3 (0.12)2 = 0.00205 kg-m2 .

Therefore,

I(z

axis)

= I(z

axis)1

+ I(z

axis)2

− I(z

axis)3

= 0.00911 kg-m2 .


1

Problem 18.121

Determine the moment of inerti a of the machine part in Problem 18.120 about the x axis. Solution: We divide the machine part into the 3 parts shown in the solution to Problem 18.120. Using Appendix C and the parallel axis theorem, the moments of inertia of the parts about the x axis are:

I(x

axis)1

=

1 12

m1 [(0.08)2 + (0.04)2 ]

= 0.0007168 kg-m2

I(x

axis)2

= m2



1 12

1

(0.04)2 +

4

(0.04)2



= 0.0001501 kg-m2

I(x

axis)3

= m3



1 1 2 2 12 (0.04) + 4 (0.02)



= 0.0000328 kg-m2 .

Therefore, I(x axis) = I(x

axis)1

+ I(x

axis)2

− I(x

axis)3

= 0.000834 kg-m2 .


1

Problem 18.122

The object shown consists of steel of density ρ = 7800 kg/m 3 of width w = 40 mm. Determine the moment of inertia about the axis L0 .

20 mm

O

100 mm

Solution: Divide the object into four parts:

Part (1): Part (2): Part (3): Part (4): Part (1)

The semi-cylinder of radius R = 0.02 m, height h1 = 0.01 m. The rectangular solid L = 0.1 m by h 2 = 0.01 m by w = 0.04 m. The semi-cylinder of radius R = 0.02 m, h 1 = 0.01 m The cylinder of radius R = 0.02 m, height h = 0.03 m.

ρπ R2 h1 m1 = = 0.049 kg , 2

I1 = Part (2)

10 mm

30 mm L

Part (4)

0

m4 = ρπ R2 h = 0.294 kg , I4 =

1 2



m4 (R2 ) + m4 L2 = 0.003 kg-m2 .

The composite:

m1 R 2 = 4.9 × 10−6 kg-m2 , 4

IL0 = I1 + I2 − I3 + I4 = 0.00367 kg-m2

m2 = ρwLh2 = 0.312 kg, I2 = (1/12)m2 (L2 + w2 ) + m2 (L/2)2 = 0.00108 kg-m2 .

Part (3)

m3 = m1 = 0.049 kg,

I3 = −

2

  4R 3π



m2 + I1 + m3 L −

4R 3π

2



= 0.00041179 kg-m2 .


1

Problem 18.123

Determine the moment of inerti a of the object in Problem 18.122 about the axis through the center of mass of the object parallel to L0 . Solution: The center of mass is locat ed relative to L 0 is given by

 

m1 − x

=

4R 3π

+ m2 (0.05) − m3



0.1 −

4R 3π



+ m4 (0.1)

m1 + m2 − m3 + m4

= 0.066 m ,

Ic = −x2 m + ILo = −0.00265 + 0.00367 = 0.00102 kg-m2


1

Problem 18.124

The thic k plate consists of steel of density ρ = 15 slug/ft3 . Determine the moment of inertia of the plate about the z axis.

y

4 in

in 2

y

in 2 x

Solution: Divide the object into three parts: Part (1) the rectangle 8 in by 16 in, Parts (2) & (3) the cylindrical cut outs.

Part (1):

m1 = ρ 8(16)(4) = 4.444 slugs.

z

4 in

4 in

8 in

4 in

4 in

I1 = (1/12)m1 (162 + 82 ) = 118.52 slug-in2 .

Part (2):

m2 = ρπ (22 )(4)

I2 =

Part (3):

m 2 (2 2 )

2

  1

123

= 0.4363 slug,

+ m2 (42 ) = 7.854 slug-in2 .

m3 = m2 = 0.4363 slugs, I3 = I2 = 7.854 slug-in2 .

The composite: Iz-axis = I1 − 2I2 = 102.81 slug-in2 Iz-axis = 0.715 slug-ft2


1

Problem 18.125

Determine the moment of inerti a of the object in Problem 18.124 about the x axis. Solution: Use the same divisions of the object as in Problem 18.124.

Part (1) : I 1x -axis =

  1

12

m1 (82 + 42 ) = 29.63 slug-in2 ,

Part (2) : I 2x -axis = (1/12)m2 (3(22 ) + 42 ) = 1.018 slug-in2 . The composite: Ix -axis = I1x -axis − 2I2x -axis = 27.59 slug in 2

= 0.1916 slug ft 2


1

Problem 18.126

The airplane is at the beginning of its takeoff run. Its weight is 1000 lb. and the initial thrust T exerted by its engine is 300 lb. Assume that the thrust is horizontal, and neglect the tangential forces exerted on its wheels. (a)

T

If the acceleration of the airplane remains constant, how long will it take to reach its takeoff speed of 80 mi/hr? Determine the normal force exerted on the forward landing gear at the beginning of the takeoff run.

(b)

6 in

ft 1

ft 7

Solution: The acceleration under constant thrust is

a=

T m

=

300(32.2) 1000

W

= 9.66 ft/s2 . T

The time required to reach 80 mph = 117.33 ft/s is

t=

v a

=

117.33

= 12.1 s

9.66

The sum of the vertical forces: the moments:





Fy = R + F − W = 0. The sum of

R

F

M = 7F − 0.5T − 1R = 0. Solve: R = 856.25 lb,

F = 143.75 lb


1

Problem 18.127

The pulleys can turn fre ely on their pin supports. Their moments of inertia are IA = 2 0.002 kg-m , IB = 0.036 kg-m 2 , and IC = 0.032 kg-m 2 . They are initially stationary, and at t = 0 a constant M = 2 N-m is applied at pulley A . What is the angular velocity of pulley C and how many revolutions has it turned at t = 2 s?

100 mm 100 mm

A

Solution: Denote the upper and lower belts by the subscripts U and L . Denote the difference in the tangential component of the tension in the belts by

C

TUB TUA B

RA TLA

TB = TLB − TUB .

B

RB 1

A

TA = TLA − TUA ,

200 mm 200 mm

C

RC

TLB RB 2

From the equation of angular motion: M + RA TA = IA αA , −RB TA + RB TB = IB αB , −RC TB = IC αC . From kinematics, RA αA1 = RB 1 αB ,2R B 2 αB = RC αC , from which αA =

αB =

RB 1 RC RA RB 2 RC RB 2

αC =

αC =

0.2 0.1

(0.2)(0.2) (0.1)(0.1)

αC = 4αC ,

αC = 2αC .

Substitute and solve: α C = 38.5 rad/s2 , from which ωC = αC t = 76.9 rad/s

N=θ

  1

2π

=

αC

4π

(22 ) = 12.2 revolutions


1

Problem 18.128

A 2-kg box is subj ected to a 40-N horizontal force. Neglect friction. (a) (b)

40 N

c

If the box remains on the floor, what is its acceleration? Determine the range of values of c for which the box will remain on the floor when the force is applied.

A

B

100 mm

100 mm

Solution:

(a)

From Newton’s second law, 40 = (2)a , from which

40 N C

a = 40 = 20 m/s2 . 2

(b)

mg



Fy = A + B − mg = 0. The sum of The sum of forces:  the moments about the center of mass: M = 0.1B − 0.1A − 40c = 0. Substitute the value of B from the first equation into the second equation and solve for c :

c=

A

B

100 mm

100 mm

(0.1)mg − (0.2)A

40

The box leg at A will leave the floor as A ≤ 0, from which

c≤

(0.1)(2)(9.81)

40

≤ 0.0491 m

for values of A ≥ 0.


1

Problem 18.129

The slen der, 2-slug bar AB is 3 ft long. It is pinned to the cart at A and leans against it at B . (a) (b)

If the acceleration of the cart is a = 20 ft/s 2 , what normal force is exerted on the bar by the cart at B ? What is the largest acceleration a for which the bar will remain in contact with the surface at B ?

B a

A

60°

Solution: Newton’s second law applied to the cente r of mass of the bar yields

B

− B + Ax = maGx , Ay − W = maGy ,

− Ay



L cos θ 2



+ (B + Ax )



L sin θ 2



= IG α,

where aGx , aGy are the accelerations of the center of mass. From kinematics, aG

W

Ax Ay

2 2 = aA + α × rG/A − ωAB rG/A = 20i ft/s

where α = 0, ωAB = 0 so long as the bar is resting on the cart at B and is pinned at A . Substitute the kinematic relations to obtain three equations in three unknowns: − B + Ax = ma,A y − W = 0,

− Ay



L cos θ

2



+ (B + Ax )



L sin θ

2



= 0.

W cot θ

ma − . For W = mg = 64.34 lb, θ = 60◦ , m = 2 2 2 2 slug, and a = 20 ft/s , B = −1.43 lb, from which the bar has moved away from the cart at point B. (b) The acceleration that produces a zero normal force is

Solve: B =

a = g cot θ = 18.57 ft/s2 .


1

Problem 18.130

To determine a 4.5-kg tire’s moment of inertia, an engineer lets the tire roll down an inclined surface. If it takes the tire 3.5 s to start from rest and roll 3 m down the surface, what is the tire’s moment of inertia about its center of mass?

330 mm

15˚

Solution: From Newton’s second law and the angular equation of motion,

α

R

mg sin15 ◦ − f = ma, Rf = I α.

mg

a

f

From these equations and the relation a = Rα , we obtain

a=

mg sin15 ◦ . m + I /R 2

N (1)

We can determine the acceleration from

s=

3=

1 2 1 2

at 2 :

a( 3.5)2 ,

obtaining a = 0.490 m/s2 . Then from Eq. (1) we obtain

I = 2.05 kg-m2 .


1

Problem 18.131 Pulley A weighs 4 lb, IA = 0.060 slug-ft2 , and IB = 0.014 slug-ft2 . If the system is released from rest, what distance does the 16-lb weight fall in 0.5 s?

8 in

B

A

12 in

16 lb

8 lb

Solution: The strategy is to apply Newto n’s second law and the equation of angular motion to the free body diagrams. Denote the rightmost weight by W R = 16 lb, the mass by m R = 0.4974 slug, and the leftmost weight by WL = 4 + 8 = 12 lb, and the mass by mL = 0.3730 slug. R B = 8 in. is the radius of pulley B, I B = 0.014 slug-ft2 , and RA = 12 in. is the radius of pulley A, and IA = 0.060 slug-ft2 . Choose a coordinate system with the y axis positive upward.

T3

T2

T2

T1 T1

The 16 lb. weight: (1) T 1 − WR = mR aRy . WL

Pulley B: The center of the pulley is constrained against motion, and the accelera tion of the rope is equal (except for direction) on each side of the pulley. (2) − RB T1 + RB T2 = IB αB . From kinematics, (3) aRy = RB αB . Combine (1), (2) and (3) and reduce:

(4) T2 = WR +



IB 2 RB

+ mR



WR

The total system: Equate (4) and (9) (the two expressions for solve:



aRy aRy =

Pulley A: (5) T2 + T3 − WL = mL aAy , where aAy is the acceleration of the cent er of the pulley. (6) −RA T3 + RA T2 = IA αA . From the kinematics of pulley A, the acceleration of the left side of the pulley is zero, so that the acceleration of the right side relative to the left



+

IB RB2

WL 2

− WR

+ mR +



IA 2 4RA

+

mL 4



T 2 ) and

.

Substitute numerical values: a Ry = −15.7 ft/s 2 . The distance that the 16 lb. weight will fall in one-half second is

side is

s=

aright = −aRy j = aleft + αA × (2RA i)

=

 

i 0 R 2 A

j 0 0

k αA 0

 

aRy 2

t2 =

−15.7

8

= −1.96 ft

= 0 + 2RA αA j,

from which (7) a Ry = −2RA αA , where the change in direction of the acceleration of the 16 lb. weight across pulley B is taken into account. Similarly, the acceleration of the right side relative to the acceleration of the center of the pulley is aAright = −aRy j = aA + αA × (RA i) = aA + RA αA j,

from which (8) a Ay = − to obtain (9) T 2 =

WA 2

aRy

−

2



. Combine (5), (6), (7) and (8) and reduce

IA 2 4 RA

+

mA 4



ay .


1

Problem 18.132

Model the excavator’s arm ABC as a single rigid body. Its mass is 1200 kg, and the moment of inertia about its center of mass is I = 3600 kg-m2 . If point A is stationary, the angular velocity of the arm is zero, and the angular acceleration is 1.0 rad/s 2 counterclockwise, what force does the vertical hydraulic cylinder exert on the arm at B ?

y C

B 3.0 m 2.4 m

A

x

1.7m

1.7m

Solution: The distance from A to the center of mass is d=



(3.4)2 + (3)2 = 4.53 m.

The moment of inertia about A is IA = I + d 2 m = 28,270 kg-m2 .

B

Ax

mg

Ay

From the equation of angular motion: 1.7B − 3.4mg = IA α . Substitute α = 1.0 rad/s2 , to obtain B = 40,170 N.


1

Problem 18.133 Model the excavator’s arm AB C as a single rigid body. Its mass is 1200 kg, and the moment of inertia about its center of mass is I = 3600 kg-m2 . The angular velocity of the arm is 2 rad/s counterclockwise and its angular acceleration is 1 rad/s2 counterclockwise. What are the components of the force exerted on the arm at A ? Solution: The acceleration of the cente r of mass is

aG =

α

× rG/A − ω 2 rG/B =

 

i 0 3.4

j 0 3

k α 0

  − ω (3.4i + 3j) 2

= −16.6i − 8.6j m/s 2 .

From Newton’s second law: Ax = maGx = −19,900 N , Ay + B − mg = maGy .

From the solution to Problem 18.132, B = 40,170 N, from which Ay = −38,720 N


1

Problem 18.134 To decrease the angle of elevation of the stationary 200-kg ladder, the gears that raised it are disengaged, and a fraction of a second late r a second set of gears that lower it are enga ged. At the inst ant the gears that raised the ladder are disengaged, what is the ladder’s angular acceleration and what are the components of force exerted on the ladde r by its suppo rt at O ? The moment of inertia of the ladder about O is I0 = 14,000 kg-m 2 , and the coordinates of its center of mass at the instant the gears are disengaged are x = 3 m, y = 4 m.

y

O

Solution: The moment about O, − mgx = Io α , from which α=−

(200)(9.81)(3)

14,000

x

y (x , y)

= −0.420 rad/s2 .

mg Fy

The acceleration of the center of mass is

aG =

α

× rG/O −

ω2 r

G/O

i = 0 3

j 0 4

x

k α 0

  = −4αi + 3α j

Fx

2

aG = 1.68i − 1.26j ( m/s ).

From Newton’s second law: F x = maGx = 336 N, F y − mg = maGy , from which F y = 1710 N


1

Problem 18.135 The slender bars each weigh 4 lb and are 10 in. long. The homogenous plate weighs 10 lb. If the system is released from rest in the position shown, what is the angular acceleration of the bars at that instant?

45° 8 in

40 in

Solution: From geometry, the system is a parallelogram, so that the plate translates without rotating, so that the acceleration of every point on the plate is the same. Newton’s second law and the equation of angular motion applied to the plate: −FAx − FBx = mp aP Gx , F Ay + FBy − Wp = mp aP Gy . The motion about the center of mass:

Ay FAx

4 lb.

Ax

By

FAy

− FAy

  20 12

20 12

 

+ FBy

  4

+ FAx

12

+ FBx

FAy

  4

FAx

5

12

FBx

  5

12

  5

cos θ + FAy

cos θ + Ay

  5

12

  5

12

sin θ − Ax

 

  5

12

sin θ − Bx

  5

From kinematics: the acceleration of the center of mass of the bars in terms of the acceleration at point A is

aBG =

=

α

× rG/A − ω2 rG/A =

5 12

sin θα i −

5 12

i 0 −

5

j 0

cos θ

12

−

5 12

k α

sin θ

0

  

cos α j ( ft/s2 ).

From which aBGx =

  5

12

sin θ α,

a BGy = −

  10

(1)

−FAx − FBx =

(2)

FAy + FBy − Wp = −

(3)

−20FAy + 4FAx + 20FBy + 4FBx = 0,

(4)

−FAy + Ay − WB = −

12

mp sin θα ,

  5

12

 

12

sin θ = IB α.

  

Substitute to obtain the nine equations in nine unknowns:

5

12

sin θ = IB α,

12

cos θ + FBy

cos θ + By

FBx

= Ip α = 0.

Newton’s second law for the bars: −FAy + Ay − WB = mB aBGy , FAx + Ax = mB aBGx . −FBy + By − WB = mB aBGy . FBx + Bx = mB aBGx . The angular acceleration about the center of mass:

 

FBy FBy

FAx

12

Bx

FBx 4 lb.

cos θα,

  5

10 12

mp cos θ α ,

  5

12

mB cos θ α ,

(5)

FAx + Ax =

(6)

FAx sin θ + FAy cos θ − Ax sin θ + Ay cos θ =

(7)

FBx + Bx =

(8)

−FBy + By − WB = −

(9)

FBx cos θ + FBy sin θ − Bx cos θ + By sin θ =

12

5 12

 

mB sin θα ,

  12 5

IB α ,

mB sin θα ,

  5

12

mB cos θα ,

  12

IB α . The 5 number of equations and number of unknowns can be reduced by combining equations, but here the choice is to solve the system by iteration using TK Solver Plus . The results: F Ax = −2.21 lb, FAy = 1.68 lb, FBx = −3.32 lb, Ax = 3.32 lb, Ay = 4.58 lb, Bx = 4.42 lb, B y = 5.68 lb. α = 30.17 rad/s2 .

since ω = 0 upon release. The acceleration of the plate:

aP = α × rP /A − ω 2 rP /A =

  

i 0 10 cos θ − 12

j 0 10 sin θ − 12

k α

0

  

10 10 2 = 12 sin θα i − 12 cos θα j ( ft/s ). From which

aP x =

  10 12

sin θα,a Py = −

  10 12

cos θα.


1

Problem 18.136 A slender bar of mass m is released from rest in the position shown. The static and kinetic friction coefficients of friction at the floor and the wall have the same value µ . If the bar slips, what is its angular acceleration at the instant of release? Solution: Choose a coordinate system with the origin at the intersection of wall and floor, with the x axis parallel to the floor. Denote the points of contact at wall and floor by P and N respectively, and the center of mass of the bar by G . The vector locations are rN = iL sin θ, rP = jL cos θ, rG =

L

2

l

(i sin θ + j cos θ) . θ

From Newton’s second law: P − µN = maGx , N + µP − mg = maGy ,

where aGx , aGy are the accel erations of the cente r of mass. The moment about the center of mass is Substitute to obtain the three equations in three unknowns,

MG = rP /G × (P i + µP j) + rN/ G × (Nj − µNi) :

MG =

PL 2

MG = −

 

i − sin θ 1

  PL 2

j cos θ µ

k 0 0

 

 

NL

+

2

i sin θ −µ

  NL

(cos θ + µ sin θ) k +

2

j − cos θ 1

k 0 0

 

.

P − µN =

(2)

µP + N = −

(3)

α, 2 mL sin θ α + mg. 2 PL NL (cos θ + µ sin θ ) + (sin θ − µ cos θ) = IB α . − 2 2

(sin θ − µ cos θ )k

Solve the first two equations for P and N :

From the equation of angular motion, P =

−

  PL 2

(cos θ + µ sin θ ) +

  NL

mL

2(1 + µ2 )

2

N =−

α

= aN i +

aG =



aN −

from which

× rG/N − ω 2 rG/N

  

−

j 0 L cos θ

2

2

αL cos θ

2 aGy = −

 

i+ −

L sin θ

2

mL

2(1 + µ2 )

µmg (1 + µ2 )

.

k α

0

αL sin θ

2

mL2

4



  

1 − µ2 1 + µ2

 

sin θ − mgL

Substitute I B =



(sin θ + µ cos θ)α +

mg (1 + µ2 )

.

Substitute the first two equations into the third, and reduce to obtain

α IB +

i 0 L sin θ

(cos θ − µ sin θ)α +

(sin θ − µ cos θ) = IB α

From kinematics: Assume that at the instant of slip the angular velocity ω = 0. The acceleration of the center of mass in terms of the acceleration at point N is aG = aN +

mL cos θ

(1)

j, α=

µ

=

 

1 + µ2 1 12

 

1 − µ2

mgL

2



1 + µ2



cos θ.

mL2 , reduce, and solve:

(3(1 − µ2 ) sin θ − 6µ cos θ)g (2 − µ2 )L

α.

The acceleration of the center of mass in terms of the acceleration at point P is a G = aP + α × rG/P .

P

µP

aG = aP + α × rG/P − ω 2 rG/P

θ = aP j +

aG =



  

i 0 L sin θ

αL cos θ 2

from which

−

2

  i+

aGx =

j 0 L cos θ

2

k α

0

αL sin θ aP + 2

L cos θ

2

   

mg

µN

,

N j,

α.


1

Problem 18.137

Each of the go-cart’s fron t wheels weighs 5 lb and has a moment of inertia of 0.01 slug-ft2 . The two rear wheels and rear axle form a single rigid body weighing 40 lb and having a moment of inertia of 0.1 slug-ft2 . The total weight of the go-cart and driver is 240 lb. (The location of the center of mass of the gocart and driver, not including the front wheels or the rear wheels and rear axle, is shown.) If the engine exerts a torque of 12 ft-lb on the rear axle, what is the go-cart’s acceleration?

15 in 6 in

4 in

A

B 16 in 60 in

Solution: Let a be the cart’s acceler ation and αA and αB the wheels’ angular accelerations. Note that

12 ft-lb

A a = (6/12)αA , a = (4/12)αB .

(2)

x

Ay

12 ft-lb Ay

Front wheel:



40 lb Fx = Bx + fB = (10/32.2)a,

B

x

(1)

(3)



Fy = By + NB − 10 = 0,

(4)



M = −fB (4/12) = (0.02)αB .

(5)

Ax

240 – 50 lb

By By 10 lb

NA

Bx fB

fA NB

Rear Wheel:



Fx = Ax + fA = (40/32.2)a,

(6)



Fy = Ay + NA − 40 = 0,

(7)



M = 12 − fA (6/12) = (0.1)αA .

(8)

Cart:



Fx = −Ax − Bx = (190/32.2)a,

(9)



Fy = −Ay − By − 190 = 0,

(10)



M = Bx [(15 − 4)/12] + By [(60 − 16)/12] + Ax [(15 − 6)/12] − Ay (16/12) − 12 = 0.

(11)

Solving Eqs. (1)–(11), we obtain a = 2.99 ft/s2 .


1

Problem 18.138 Bar AB rotates with a constant angular velocity of 10 rad/s in the counterclockwise direction. The masses of the slender bars BC and CD E are 2 kg and 3.6 kg, respectively. The y axis points upward. Determine the components of the forces exerted on bar BC by the pins at B and C at the instant shown.

y B

400 mm

C

10 rad/s

A

D

x

400 mm

700 mm

Solution: The velocity of point

vB = ωAB × rB =

 

i 0 0

j

k

0 10 0.4 0

B is

E

700 mm

By Bx

 

Cy C

= −0.4(10)i = −4i ( m/s).

x

WBC Cx

The velocity of point C is

Cy

vC = vB +

ωBC

× rC/B = −4i +

 

i 0 0.7

j 0 −0.4

k ωBC 0

 

= −4i + 0.4ωBC i + 0.7ωBC j ( m/s).

vC = vD +

=0+

ωCDE

 

j 0 0

 

k ωCDE 0

 

i 0 0.35

j 0 −0.2

k αBC 0

 

2 (0.35i − 0.2j), − ωBC

from which a G = 61.25i + 148.44j ( m/s2 ) The equations of motion: Bx + Cx = mBC aGx , By + Cy − mBC g = mBC aGy , where the accelerations aGx , aGy are known. The moment equation, 0.35Cy + 0.2Cx − 0.2Bx − 0.35By = IBC αBC , where α BC , is known, and

× rC/D

i 0 −0.4

The acceleration of the center of mass of B C is

aG = −40j +

From the constraint on the motion at point C , vC = vC j. Equate components: 0 = −4 + 0.4ωBC , vC = 0.7ωBC , from which ωBC = 10 rad/s, v C = 7 m/s. The velocity at C in terms of the angular velocity ωCDE ,

WCE

= −0.4ωCD E j,

7 from which ω CDE = − 0.4 = −17.5 rad/s.

IBC =

 

where

ID =

1

12

mBC L2BC = 0.1083 kg-m2 , 0.4Cy − 0.15mCE g = ID αCE ,

1 12

 

mCE L2CE + (0.15)2 mCE = 0.444 kg-m2 ,

The acceleration of point B is 2 2 r 2 aB = −ωAB B = −(10 )(0.4) j = −40j ( m/s ).

The acceleration at point C is a C = aB +

aC = −40j +

 

i 0 0.7

j 0 −0.4

k αBC 0

 

αBC

2 rC/B . × rC/B − ωBC

2 (0.7i − 0.4j) (m/s2 ). − ωBC

2 2 aC = +(0.4αBC − 0.7ωBC )i + (−40 + 0.7αBC + 0.4ωBC ) j ( m/s2 ).

is the moment of inertia about the pivot point D , and 0.15 m is the distance between the point D and the center of mass of bar CDE . Solve these four equations in four unknowns by iteration: Bx = −1959 N, By = 1238 N, Cx = 2081 N, Cy = −922 N .

The acceleration in terms of the acceleration at D is

aC =

 

i 0 −0.4

j 0 0

k αCD E 0

 

2 (−0.4i) − ωCDE

= −0.4αCDE j + 0.4ω 2

i.

CDE

Equate components and solve: αBC = 481.25 rad/s2 , αCD E = −842.19 rad/s2 .


1

Problem 18.139 At the instant shown, the arms of t he robotic manipulator have the constant counterclockwise angular velocities ωAB = −0.5 rad/s, ωBC = 2 rad/s, and ω CD = 4 rad/s. The mass of arm C D is 10 kg, and the center of mass is at its midpoint. At this instant, what force and couple are exerted on arm CD at C ?

y 250

m 0m 30 30°

B

mm

20°

D C

A

Solution: The relative vector locations of ◦

B , C , and D are

Cy

◦

Cx

rB/A = 0.3(i cos30 + j sin30 )

C

D 125 mm

= 0.2598i + 0.150j (m), ◦

x 250 mm

mg

◦

rC/B = 0.25(i cos20 − j sin20 ) = 0.2349i − 0.08551j ( m), rD/C = 0.25i ( m).

The acceleration of point B is ◦

◦

2 aB = −ωAB rB/A = −(0.52 )(0.3cos30 i + 0.3sin30 j),

aB = −0.065i − 0.0375j (m/s2 ).

The acceleration at point C is 2 2 aC = aB − ωBC rC/B = aB − ωBC (0.2349i − 0.08551j).

aC = −1.005i + 0.3045j (m/s2 ).

The acceleration of the center of mass of C D is 2 aG = aC − ωCD (0.125i) (m/s2 ),

from which aG = −3.005i + 0.3045j ( m/s2 ).

For the arm C D the three equations of motion in three unknowns are Cy − mCD g = mCD aGy , Cx = mCD aGx , M − 0.125Cy = 0,

which have the direct solution: Cy = 101.15 N, Cx = −30.05 N. M = 12.64 N-m ,

where the negative sign means a direction opposite to that shown in the free body diagram.


1

Problem 18.140 Each bar is 1 m in leng th and has a mass of 4 kg. The inclined surface is smooth. If the system is released from rest in the position shown, what are the angular accelerations of the bars at that instant?

A

45°

O

◦

Solution: For convenience, denote

B 30°

◦

θ = 45 , β = 30 , and L =

Ay Ax

1 m. The acceleration of point A is

aA =

αOA

× rA/O =

 

i 0 L cos θ

j 0 L sin θ

k αOA 0

 

.

Ax Ay

mg

mg

The acceleration of A is also given by aA = aB +

αAB

aA = aB +

 

The equations of motion for the bars: for the pin supported left bar:

× rA/B . (5) Ay L cos θ − Ax L sin θ − mg

i 0 −L cos θ

j 0 L sin θ

k αAB 0

 

.

mL2

 

where I OA =

3

=

4 3

  L

2

cos θ = IOA αOA ,

kg-m2 .

aA = aB − iαAB L sin θ − jαAB L cos θ (m/s2 ).

The equations of motion for the right bar:

From the constraint on the motion at B , Equate the expressions for the acceleration of A to obtain the two equations:

(6) − Ax − B sin β = maGABx , (7) − Ay − mg + B cos β = maGABy ,

(1) − αOA L sin θ = aB cos β − αAB L sin θ,

and ( 2) αOA L cos θ = aB sin β − αAB L cos θ.

(8 ) A y

The acceleration of the center of mass of AB is

= aA +

aGAB = aA +

i 0 L cos θ

2

LαAB

2

−

j 0 L sin θ

sin θ i +

2 αAB L

2

k αAB

0

  

  L

2

cos θ + Ax

  L

2

sin θ + B

  L

2

sin θ cos β

L

−B

aGAB = aA + αAB × rGAB/A

  

B

30°

aA = αOA (−iL sin θ + jL cos θ) (m/s2 ).

  2

cos θ sin β = ICAB αAB ,

where I GAB =

  1

12

mL2 =

 1 3

kg-m2 .

,

These eight equations in eight unknowns are solved by iteration: A x = −19.27 N, Ay = 1.15 N, αOA = 0.425 rad/s2 , αAB = −1.59 rad/s2 , B = 45.43 N, a GABx = −0.8610 m/s2 , a GABy = −0.2601 m/s2

cos θ j (m/s2 ),

from which (3) aGABx = −αOA L sin θ +

(4) aGABy = αOA L cos θ +

LαAB

2

LαAB

2

sin θ (m/s2 ),

cos θ.


1

Problem 18.141 Each bar is 1 m in leng th and has a mass of 4 kg. The inclined surface is smooth. If the system is released from rest in the position shown, what is the magnitude of the force exerted on bar OA by the support at O at that instant? Solution: The acceleration of the center of mass of the bar O A is

aGOA =

αOA

aGOA = −

× rG/A = aA

L sin θ

2

αOA i +

  + 

i 0 L cos θ

j 0 L sin θ

2

2

L cos θ

2

αOA j ( m/s2 ).

k αOA

0

  , 

Ay Ax Fy mg Fx

The equations of motion:

Use the solution to Problem 18.140: θ = 45◦ , αGA = 0.425 rad/s2 , Ax = −19.27 N, m = 4 kg, from which Fx = 18.67 N, Fy =

Fx + Ax = maGOAx , Fy + Ay − mg = maGOAy .

38.69 N, from which |F| =

F

2 x

+ Fy2 = 42.96 N


1

Problem 18.142

The fixed ring gear lies in the horizontal plane. The hub and planet gears are bonded together. The mass and moment of inertia of the combined hub and planet gears are mHP = 130 kg and IHP = 130 kg-m 2 . The moment of inertia of the sun gear is Is = 60 kg-m 2 . The mass of the conn ecting rod is 5 kg, and it can be modeled as a slender bar. If a 1 kNm counterclockwise couple is applied to the sun gear, what is the resulting angular acceleration of the bonded hub and planet gears?

Planet gear Hub gear 140 mm

Connecting rod

340 mm 240 mm 720 mm

Sun gear Ring gear

Solution: The moment equation for the sun gear is

(1) M − 0.24F = Is αs . Hub & Planet Gears

For the hub and planet gears:

(2) (0.48)αHP = −0.24αs , (3) F − Q − R = mHP (0.14)(−αHP ), (4) (0.14)Q + 0.34F − IHP (−αHP ).

R

Q R

F F M S u nG e a r

Co n n ec t i n gR o d

For the connecting rod:

(5) (0.58)R = ICR αCR ,

where I CR =

 1 3

mGR (0.582 ) = 0.561 kg-m2 .

(6) (0.58)αCR = −(0.14)αHP .

These six equations in six unknowns are solved by iteration: F = 1482.7 N, αs = 10.74 rad/s2 , αHP = −5.37 rad/s2 , Q = 1383.7 N, R = 1.25 N, αCR = 1.296 rad/s2 .


1

Problem 18.143 The system is stationary at the instant shown. The net force exerted on the piston by the exploding fuel-air mixture and friction is 5 kN to the left. A clockwise couple M = 200 N-m acts on the crank AB . The moment of inert ia of the crank about A is 0.0003 kg-m2 . The mass of the connecting rod BC is 0.36 kg, and its center of mass is 40 mm from B on the line from B to C . The connecting rod’s moment of inertia about its center of mass is 0.0004 kg-m 2 . The mass of the piston is 4.6 kg. What is the piston’s acceleration? (Neglect the gravitational forces on the crank and connecting rod.) Solution: From the law of sines: sin β 0.05

=

sin40 ◦ 0.125

125 m 50

m

m m

B 40°

A

C

M

from which

(3) aGCRx = aC − 0.085αBC sin β (m/s2 ),

,

(4) aGCRy = −0.085αBC cos β (m/s2 ).

from which β = 14.9◦ . The vectors ◦

The equations of motion for the crank :

◦

rB/A = 0.05(i cos40 + j sin40 )rB/A

(5) By (0.05cos40 ◦ ) − Bx (0.05sin40 ◦ ) − M = IA αAB

= 0.0383i + 0.0321j ( m).

For the connecting rod : rB/C = 0.125(−i cos β + j sin β) (m).

(6) − Bx + Cx = mCR aGCRx

rB/C = −0.121i + 0.0321, (m).

(7) − By + Cy = mCR aGCRy

The acceleration of point B is

(8) Cy (0.085cos β) + Cx (0.085sin β) + Bx (0.04sin β)

2 r aB = αAB × rB/A − ωAB B/A ,

aB =

 

i 0 0.0383

j 0 0 .0321

k αAB 0

+ By (0.04cos β) = IGCR αBC

 

For the piston : (9) − Cx − 5000 = mP aC .

2 (0.0383i + 0.0321j) (m/s2 ). − ωAB

These nine equations in nine unknowns are solved by iteration: The acceleration of point B in terms of the acceleration of point C is

aB = aC +

αBC

× rB/C = aC

 i+

i j 0 0 −0.121 0 .0321

k

αBC 0

 

αAB = 1255.7 rad/s2 , αBC = −398.2 rad/s2 , aGCRx = −44.45 m/s 2 , aGCRy = 32.71 m/s 2 , By = 1254.6 N, Bx = −4739.5 N,

2 (−0.121i + 0.0321j) (m/s2 ). − ωBC

Cx = −4755.5 N, Cy = 1266.3 N,

Equate the two expressions for the acceleration of point B , note ωAB = ωBC = 0, and separate components:

aC = −53.15 m/s 2 .

(1) − 0.05αAB sin40 ◦ = aC − 0.125αBC sin β,

B

(2) 0 .05αAB cos40 ◦ = −0.125αBC cos β. The acceleration of the center of mass of the connecting rod is aGCR = aC +

αBC

aGCR = aC i +



A

2 rGCR/C , × rGCR/C − ωBC

i 0

 −0.085cos β

j 0

k αBC



0.085sin β

0



2 ( − ωBC −0.085cos β i + 0.085sin β j) (m/s2 ),

Bx

0.05

β

40

Cx

Cy

G

0.125

C

Cx

By

By

5000 N

Bx Ay

Cy

N

Ax

M


1

Problem 18.144 If the crank AB in Problem 18.143 has a counterclockwise angular velocity of 2000 rpm at the instant shown, what is the piston’s acceleration? Solution: The angular velocity of AB is ωAB = 2000

  2π

60

= 209.44 rad/s.

The angular velocity of the connecting rod BC is obtained from the expressions for the velocity at point B and the known value of ω AB :

vB = ωAB × rB/A =

 

i 0 0.05cos40

j 0 0.05sin40

◦

◦

k ωAB 0

◦

 

.

◦

vB = −0.05sin40 ωAB i + 0.05cos40 ωAB j ( m/s).

vB = vC i +

 

i 0 −0.125cos β

j 0 0.125sin β

k ωBC 0

 

,

vB = vC i − 0.125sin βω BC i − 0.125cos βω BC j ( m/s). ◦

From the j component, 0 .05cos40 ωAB = −0.125cos βω BC , from which ωBC = −66.4 rad/s. The nine equations in nine unknowns obtained in the solution to Problem 18.143 are 2 (1) − 0.05αAB sin40 ◦ − 0.05ωAB cos40 ◦ 2 = aC − 0.125αBC sin β + 0.125ωBC cos β, 2 (2) 0 .05αAB cos40 ◦ − 0.05ωAB sin40 ◦ 2 = −0.125αBC cos β − 0.125ωBC sin β, 2 (3) aGCRx = aC − 0.085αBC sin β + 0.085ωBC cos β (m/s2 ),

(4) a

GCRy

= −0.085α

BC

cos β − 0.085ω 2

BC

sin β (m/s2 ),

(5) By (0.05cos40 ◦ ) − Bx (0.05sin40 ◦ ) − M = IA αAB , (6) − Bx + Cx = mCR aGCRx , (7) − By + Cy = mCR aGCRy , (8) Cy (0.085cos β) + Cx (0.085sin β) + Bx (0.04sin β) + By (0.04cos β) = IGCR αBC .

(9) − Cx − 5000 = mP aC . These nine equations in nine unknowns are solved by iteration:

αAB = −39, 386.4 rad/s 2 αBC = 22,985.9 rad/s2 , aGCRx = −348.34 m/s 2 , aGCRy = −1984.5 m/s 2 , By = 1626.7 N, Bx = −3916.7 N, Cx = −4042.1 N, Cy = 912.25 N, ac = −208.25 ( m/s2 )


1

Engineering.mechanics.dynamic.bedford.ch18

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